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THE EASIEST WAY TO UNDERSTAND ALGEBRA
Roy Richard Sawyer
APPENDIX 1. ANSWERS TO ALL EQUATIONS
APPENDIX 2. SOLUTIONS FOR ALL EQUATIONS
© Copyright 2017 by Roy Richard Sawyer  All rights reserved
INTRODUCTION
There is a voice inside you that always says," You can't do that. Leave it."
This is not true, but you believe this voice because you get some benefit from accepting the advice. You get a good excuse to slip away from your work. You can go out with friends, watch TV, or do whatever you want. However, you cannot feel genuine pleasure from your entertainment because there is a part of you who wants to be proud of your achievements. This voice tells to "Keep trying! You can do that! You are smart." Believe in what this voice is saying! This tutorial will show you a way of thinking that will help you to understand math. This tutorial is for anyone who wants to feel comfortable using a mathematical formula; who wants to comprehend the beauty of algebraic expressions.
Did you ever feel frustration looking at your math text book? Forget it! Fall in love with math!
1. HOW TO SOLVE THE EQUATIONS
Math is a very enjoyable field of activity. Having just a pen and a piece of paper you can invent whatever you want. You can wander around the paper with numbers and symbols caring about just one thing: equality should be equality, nothing more. Let us imagine that you are the first great mathematician. People are only familiar with arithmetic: how to add, subtract, multiply and divide. In school, they study boring things such as these expressions:
2 + 3 = 5 or 7  4 = 3
You are the first who suspects there is a way to express a common idea of the equations written above.
First you write: a + b = c or c  a = b. Now you have discovered common rules that can help people to solve any equation. To verify the discovery, you have to perform experiments with numbers.
Let us write a simple equation: 4 + 8 = 12
Let's add any number to the left side of the equation.
4 + 8 + 3 = 12
What did you get?
15 = 12
This is incorrect! How can you fix your equation? You must add the same number to the right side of the equation.
4 + 8 + 3 = 12 + 3. What did you get? 15 = 15
You discovered the first rule for equations. This rule says: "If you add the same number to the left side and to the right side of an equation, this equation will still be true." To express this rule in a common way you can write:
If a + b = c then a + b + n = C + n where a, b, c, n equal any numbers.
Are you a genius? Of course, you are! Go ahead. Let us try another experiment.
What happens if you subtract any number from the left side of an equation?
5+2=7
5+25=7
What did you get? 2 = 7
This is incorrect, but you know how to fix your equation. You must subtract the same number from the left and right side of the equation.
5 + 2  5 = 7  5. Then 2 = 2.
Congratulations! You discovered the second rule for equations. This rule says: "If you subtract the same numbers from the left and the right side of an equation this equation will still be true."
Or you can write:
if a + b = c then a + b  n = c  n where a, b, c, n equal any numbers.
What other kind of experiments can you do? You can multiply one side of an equation by some number. Let's write an equation:
5  1 = 4
What happens, if you multiply the left side of the equation by 7?
(5  1)7 = 4 then 28 = 4
This is not true. Try to multiply both sides of the equation by 7.
(5 1)7= 4 x 7. Then 28 = 28
You discovered one more rule for equations. The third rule says:
" If you multiply the left and the right side of an equation by the same number,
this equation will still be true."
if a  b = c then (a  b)n = (c)n
One more question. What happens if you divide one half of an equation by any number?
4 + 6 = 10
(4 + 6)/2 = 10
then 5 = 10.
You can ask yourself, "How many times will I make the same mistake?"
But you have the knowledge to fix the problem.
You must divide both sides of the equation by the same number.
(4 + 6): 2 = 10: 2 then 5 = 5
You discovered the fourth rule for equations. This rule says:
"If you divide the left and the right side of an equation by the same number the equation will still be true."
So, you can write:
if a + b = c then (a + b)/ n = c/ n
Where a, b, c, are any numbers, but n does not equal to 0 because you can't divide numbers by 0.
People will ask you, "What kind of benefit can you get from these rules?"
Your response will be, "You can use these rules to solve any equation."
Let us write an equation where one number is unknown.
X  3 = 11
How can we solve this equation? Try to apply the first rule:
If you add the same number to the left and the right side of an equation, this equation will be true.
For our equation, it is convenient to add 3 to both sides of the equation.
X  3 + 3 = 11 + 3
Since 3 + 3 = 0
Then X = 11 + 3
So, X=14
Let us try to solve an equation where all numbers are represented by lepers.
X  b = c
Apply the first rule to solve this equation
X  b + b = c + b
Since, b + b = 0, then
X = c + b.
To solve the equation X + b = c we can apply the second rule.
If X + b = c then
X + b  b = c  b
X = c  b.
The next example is X + 7 = 15
then X + 7  7 = 15  7 and X = 8
Do not proceed until you perform some exercises to become comfortable using the first few rules of equations.
Practice 1. Solving the equations.
Solve for X:
You can find the answers in appendix 1. If your answer is wrong try again.
If you can't get the right answer read the solution in appendix 2.
Let's solve the equation
4X  5 = 15
You can apply the first rule.
4X  5 + 5 = 15 + 5 then 4X = 20.
How can you find X? You can apply the fourth rule.
If you divide both sides of an equation by the same numbers, this equation will still be true.
4X / 4 = 20 / 4, then X = 5.
To solve equation
aX  b = c
Apply the first rule.
aX  b + b = c + b,
then aX = c + b
Now apply the fourth rule.
If aX = c + b, then aX / a = (c + b) / a
and X = (c + b)/ a
Do not read any more until you perform some exercises.
Practice 2. Solving the equations.
Solve for X:
Answers are in appendix 1. Solutions are in appendix 2.
If you have such an equation to solve:
X/a  5 = 6
Then apply the first rule:
X/a  5 + 5 = 6 + 5
X/a = 6 + 5
X/a = 11
Then apply the third rule.
X/a * a = 11 *a
X = 11a
Let's solve the equation:
2X  4b = 2bc
Apply the first rule:
2X  4b + 4b = 2bc + 4b,
then 2X = 2bc + 4b
Apply the third rule:
2X/ 2 = (2bc + 4b) / 2
You should know how to divide a binomial by a monomial.
If you have forgotten it, you could find the rule by yourself. Can you write?
(2bc+ 4b)/2 = 2bc/2 + 4b/2. Yes, you can.
Let' us check. Suppose, c = 2 and b = 3.
To divide a binomial by 2, try to divide each monomial by 2
2*3*2/2 + 4*3/2 = 12
Now try to solve the binomial first and then divide by 2
(2*3*2 + 4*3)/2 then 24/2 = 12
We got the same answer. It means that
(a+ b)/2 = a/2 + b/2.
We discovered a rule: To divide a binomial by a number, divide each monomial
inside the binomial by that number. Come back to your equation.
2X = 2bc + 4b. Then
2X / 2 = 2bc / 2 + 4b / 2
Then X = bc + 2b
You can factor out b and get
X = b(c + 2)
Whenever you don't know the rule, you can put any numbers in place
of the letters and check equality. Discover rules by yourself.
Let us solve a more complicated equation:
Multiply both sides of the equation by 5X.
5X  5 = 50X Use the 2nd rule, subtract 5X from both sides:
5X  5  5X = 50X  5X
or 45X =  5
Divide both sides by 45
X =  1/9
The next equation:
Find the common denominator:
Since, +aX aX = 0 our equation becomes simple:
Use the 3rd rule: multiply both sides of the equation by (a + b)
Then b  bX = c(a + b)
Apply the 2nd rule, subtract b from both sides of the equation:
b  bX  b =c (a + b)  b
Then bX =c (a + b)b Divide both sides by b:
To make this algebraic expression more beautiful, multiply the numerator and denominator by (1).
You can do that because (1):(1)= 1. If you multiply any number by 1 the number will not be changed.
then
The next equation:
2X = a  b
It is not convenient for you to have a minus in front of 2X.
You can change the equation into a more convenient form.
Let's multiply both sides of the equation by 1
(2X)(1) = (a  b)(1) then you get
2X = a + b or 2X = b  a Divide both sides by 2:
There is another way to solve this equation:
 2X = a  b
Let's divide both sides of the equation by 2
To make your result more beautiful you can multiply the numerator and the denominator by  1
The next equation:
3a  6X = 6X  9a
You can see that on the left side of the equation you have 6X
and on the right side +6X. It is more convenient for you to have a + in front of X,
therefore, you leave +6X on the right side and get rid of 6X in the left side of the equation.
Add 6X to both side of the equation:
3a  6X + 6X = 6X  9a + 6X then 3a = 12X  9a
Add 9a to both side of the equation:
3a + 9a = 12X  9a + 9a then 12a = 12X X = a
Do not read any more until you perform some exercises.
Practice 3. Solving the equations
(Solve for X)
Let us continue and discus the equation:
aX  bX = a  b
Factor out X which is a common factor for binomial aX  bX,
then you get X (a  b) = a  b
Divide each part of the equation by a  b
X(a  b)/(a  b)=(a  b)/(a  b)
X = 1
Do not read any more until you perform exercises.
Practice 4. Solving the equations.
(Solve for X)
4. aX  bX  cX = 2a  2b  2c
Let us continue and solve the equation:
aX  bX = 2b  2a Factor out X from the left side of the equation.
X(a  b) = 2b  2a Factor out 2 from the right side of the equation.
X(a  b) = 2(b  a) then divide both sides by (a  b)
Factor out (1) from the numerator
Or you can simplify this algebraic expression by factoring out (1) from the denominator
Practice 5. Solving the equations.
Solve for X.
You can find the answers in appendix 1 and the solution in appendix 2.
2. SYSTEM OF EQUATIONS
Look at the equation X + Y = 3. X and Y are unknown. You can´t find neither X nor Y from this equation. You need additional information about the "relationship" between them. Such information may be included in an additional equation. For example: X  Y =  1. Now you have a system of 2 equations:
1. X + Y = 3
X  Y = 1
There are several ways to solve it. The first way: solve for X in any equation, for example, the first one.
For that subtract Y from each side of the equation:
X + Y  Y = 3  Y; Find X
X = 3  Y
Then put (3  Y) in place of X in the second equation (X  Y = 1).
You will obtain: 3  Y  Y = 1 or 3  2Y = 1
Now solve that equation for Y. Add 2Y to both sides of the equation.
3  2Y + 2Y = 1 + 2Y
3 = 1 + 2Y
Add 1 to both sides of the equation.
3 + 1 = 1 + 1 + 2Y
4 = 2Y
Switch 4 and 2Y
2Y = 4
Divide both sides of the equation by 2.
2Y/2 = 4/2
Y = 2. Now put 2 in place of Y in any original equation to find X.
One of the original equations is X + Y = 3
X + 2 = 3;
Subtract 2 from both sides of equation.
X + 2  2 = 3  2;
X = 1
Y = 2
The second way:
1. X + Y = 3
X  Y = 1
You can sum the left sides of both equations and sum the right sides of both equations.
If X + Y = 3 and X  Y = 1 Then
(X + Y) + (X  Y) = 3 + (1) or
You can write it in this way:
Then X = 1
Put 1 in place of X in any equation.
1 + Y = 3
Subtract 1 from both sides of the equation.
1 + Y  1 = 3  1
Y = 3  1
Y = 2
We can solve system equations using their graphs. If we plot each equation, we will get two straight lines. The point of the intersection of the lines will have values X and Y that fit both equations.
To draw a graph for an equation, we have to change it to a general form: Y = aX + b
Let's start from the first equation: X + Y =3
Subtract X from both sides of the equation
X  X + Y = 3  X
Y = 3  X
Find 2 points to draw the first equation. Assign any value to X and calculate the value of Y.
X = 3; Y(3) = 3  3 =0
X = 6; Y(6) = 3  6 =  3
Two points are enough to draw a straight line.
Let's find points for the second equation. Change form to general one.
X  Y = 1
Subtract X from both sides of the equation
X  X  Y =  1  X
 Y =  1  X
Multiply both sides by 1
Y(1) = (1)(1)  X(1)
Y = 1 + X
Find 2 points to draw the second equation. Assign any value to X and calculate the value of Y.
X = 5; Y(5) = 1 + 5 = 6
X = 5; Y(5)= 1  5 = 4
Now we can draw graphs for both lines.
In graph 1, you see that the lines intersection point has X = 1 and Y =2. They are the same values we found before.
Graph 1. A point of the intersection: X=1, Y=2.
The next system of equations is:
2. 2X + Y = 5
X + Y =2
In this case, we have + in front of X and Y in both equations.
To eliminate one unknown member of the equation you have to subtract the second equation from the first one.
(2X + Y)  (X + Y) = 5  2
Then put 3 in place of X in any original equation.
3 + Y = 2
Subtract 3 from both sides of the equation.
3 + Y  3 = 2  3
Y = 2  3 =  1
Y = 1
Answers: X = 3; Y = 1;
Let 's solve these equations using graph.
The first equation is 2X + Y = 5.
Change it to a general form Y = aX + b
Subtract 2X from both sides of the equation.
2X  2X + Y = 5  2X
Y = 5  2X
Find 2 points to draw the first equation. Assign any value to X and calculate the value of Y.
X = 0; Y(0)= 5 2* 0; Y(0) = 5
X = 4; Y(4) = 5  2*4 = 5  8 =  3
Let's find 2 points for the second equation.
X + Y = 2
Change it to a general form Y = aX + b
X  X + Y = 2  X;
Y = 2  X
Find 2 points to draw the second equation. Assign any value to X and calculate the value of Y.
X = 4; Y(4) = 2  4 =  2
X =  4; Y(4) = 2  (4) = 2 + 4 = 6
In graph 2, we can see that the point of the intersection is (3, 1). The same answer we found before: X = 3, Y= 1
Graph 2.The point of the intersection: X=3, Y=1.
The next system of equation is:
3. X  Y = 3
3X  2Y = 4
To eliminate one unknown member of the equation you have to get the same number in front of X in the first and in the second equation or the same number in front of Y in the first and in the second equation. In our case, in front of Y you have 1 in the first equation and 2 in the second equation.
To get 2Y in the first equation you should multiply both sides of the equation by two. Then you will get:
2X  2Y = 6
3X  2Y = 4
Now subtract the second equation from the first one.
then X =  2
Put 2 in place of X in any original equation.
 2  Y = 3
Add 2 to both sides of the equation
 2  Y + 2= 3 + 2
 Y = 5
Multiply both sides of the equation by  1
 Y * (1) = 5 * (1)
Y =  5
You can check your result. Put 2 in place of X and 5 in place of Y in both equations:
Check the first equation:
1. X  Y = 3
2  ( 5) = 3
2 + 5 = 3
3 = 3
Check the second equation:
2. 3X  2Y = 4
3*(  2 )  2*(  5 ) = 4
 6 + 10 = 4
4 = 4
Let's solve these equations using graph.
The first equation is X  Y = 3
Change it to a general form:
X  X  Y = 3  X
 Y = 3  X multiply both sides by 1
Y = X  3
Find 2 points to draw the first equation. Assign any value to X and calculate the value of Y.
X = 0; Y(0) = 0  3 =  3
X = 3; Y(3) = 3  3 = 0
The second equation 3X  2Y = 4
Change it to a general form:
3X  3X  2Y = 4  3X
2Y = 4  3X
Divide each member of equation by 2
2Y/(2) = 4/(2)  3X/(2)
Y =  2 + 3X/2
Switch the order of the members on the right side of the equation
Y = 3X/2  2
Find 2 points to draw the second equation. Assign any value to X and calculate the value of Y.
X = 4; Y(4) = 3*4/2  2 = 6  2 = 4
X = 2; Y(2) = 3*2/2  2 = 3  2 = 1
Graph3. The point of the intersection: X=2, Y= 5.
Practice 6. Solving system of equations.
X  Y =  5
2X  2Y = 6
2X + Y =  4
X + 3Y = 0
4X + 2Y = 10
4X  2Y = 6
7X + 2Y=6
X  3Y = 3
X + 3Y= 2
X  2Y =  1
3. QUADRATIC EQUATIONS
Let us examine the following equation:
6X^2 + 3X  3 = 0
Express the left side of the equation as product of two binomials. To perform that, factor the first and the last term of the equation. Factor 6X^2.
You can express 6X^2 as 6X * X or 3X * 2X
Then factor 3. You can express 3 as 1(3) or 3(1)
6X^2 = (2X ) (3X )
Factor the last term 3 = (1) ( +3) Combine them together: You will get
6X^2 + 3X  3=(2X  1) (3X + 3)
because 6X^2 + 3X  3=2X * 3X + 2X * 3  1 * 3X 3
You may check whether the expression on the left side of the equation equals to the expression on the right side.
Is 6X^2 + 3X  3 = (2X  1) (3X + 3)?
Multiply (2X  1) by (3X + 3) and you will get
6X^2 + 6X  3X  3 or 6X^2 + 3X  3
The two expressions are equal. Now you can write
(2X  1) (3X + 3) = 0
If the product of two binomials equals 0 then each of them may be equals to 0.
(2X  1) = 0 (3X + 3) = 0
Solve the first equation:
Add 1 to both sides of the equation
2X 1 + 1=0 + 1
2X = 1 and X = 0.5
Solve the second equation: (3X + 3) = 0
Subtract 3 from both sides of the equation.
3X + 3  3=0  3
3X = 3 and X =  1
To check your solution put each result in the original equation
The original equation is: 6X^2 + 3X  3=0
6 * (0.5) (0.5) + 3(0.5)  3 = 0
1.5 + 1.5  3 = 0
3  3 = 0
You can solve the same equation in a different way.
Factor 6X^2 as 6X * X
Factor the last term and you will get
(6X  3) (X + 1) = 0
Now you can check:
(6X  3) (X + 1) = 6X^2 + 6X  3X  3=
= 6X^2 + 3X  3
You got the original equation. Let us solve it.
(6X  3) = 0
Add 3 to both sides of the equation.
6X  3 + 3 =0 + 3
6X = 3
Divide each side of the equation by 6
6X/6 = 3/6
X = 0.5
since (6X  3) (X + 1)=0
(X + 1) = 0
Solve that equation by subtracting 1 from both sides of the equation
X + 1  1 = 0 1
X = 1
You get the same results.
We can solve this equation with quadratic formula:
X =[ b +/ sqrt ( b^24ac ) ] / 2a
6X^2 + 3X  3 = 0
a = 6
b=3
c=3
X = [ b +/ sqrt ( b^24ac ) ]/2a
X= [ 3 +/ sqrt ( 3^2  4*6* (3 ) )]/2*6
X=[ 3 +/ sqrt ( 9  (72 ) ]/12
X = [ 3 +/ sqrt( 9 + 72 ) ]/12
X = [ 3 +/ sqrt( 81 ) ]/12
X = (3 +/ 9)/12
X = (3 9)/12 = 12/12 = 1
X = (3 + 9)/12 = 6/12 = 0.5
We got the same result: X = 1 and X =0.5
Practice 7. Solving Quadratic Equations
APPENDIX 1. ANSWERS TO ALL EQUATIONS
Practice 1 Answers.
1) X=5;
2) X=  8;
3) X=4+ab;
4) X= Z+Y;
5) X= C+2a;
6) X=bc3ab;
7) X=1;
8) X=a;
9) X= b;
10) X = 3a  ab or X = a(3  b);
11) X= c(2b 1);
12) X=a;
13) X=32k;
14) X=2ab;
15) X=2a;
16) X=a + c;
Practice 2 Answers.
1) X=4;
2) X=3;
3) X=6;
4) X=6;
5) X=3;
6) X=3;
7) X=a/2;
8) X=6b;
9) X=(b2a)/7;
10) X=1+a;
11) X=a;
12) X=24b;
13) X= 4b  3a;
14) X=b + 1;
15) X=1b;
16) X=c  2c/a;
Practice 3 Answers.
1) X=a4;
2) X=5/3;
3) X= (a  b)/2;
4) X=a;
5) X=2b;
6) X= bc;
7) X= b + c
Practice 4 Answers.
1) X= 2;
2) X= 1/2;
3) X= 1/(ab);
4) X=2;
5) X= 6/(bc);
6) X=c;
7) X=2a;
8) X=2;
9) X=1;
Practice 5 Answers.
1) X= 2;
2) X= 1;
3) X= 2;
4) X= 3;
5) X=  8;
6) X=  1;
7) X=2;
8) X=5/7;
Practice 6 Answers.
1) X=  2; Y=3;
2) X=2; Y=1;
3) X=3; Y=2;
4) X=3; Y=1;
5) X=2; Y=1;
6) X= 2; Y=1;
7) X=1; Y=  0.5;
8) X=3; Y=0;
9) X=1; Y=1;
10) X=1; Y=1;
Practice 7 Answers.
1. X = 5 or X = 5
2. X = 0.5 or X = 4
3. X = 1/3 or X = 2
4. X = 0.25 or X = 3
5. X = 1/7 or X = 4
6. X = 3/5 or X = 5
7. X = 1 or X = 1/3
8. X = 1 or X = 3
9. X = 3/8 or X =8/3
10. X = 0.25 or X = 4
11. X = 0.5 or X = 3
12. X = 4 or X = 0
13. X = 4.5 or X = 1
14. X = 1.5 or X = 3/7
15. X = 0.4 or X = 7
APPENDIX 2.
PRACTICE 1. SOLUTIONS
Add 5 to both sides of the equation
X  5 + 5 = 0 + 5
X = 0 + 5
X = 5
Subtract 11 from both sides of the equation :
X + 11  11 = 3  11
X = 3  11
X =  8
Add ab to both sides of the equation :
X  ab + ab = 4 + ab
X = 4 + ab
Add Y to both sides of the equation :
X  Y + Y = Z + Y
X = Z + Y
Add 2a to both sides of the equation :
X  2a + 2a = c + 2a
X = c + 2a
Subtract 3ab from both sides of the equation :
X + 3ab  3ab = bc  3ab
X = bc  3ab
Subtract k from both sides of the equation :
X + k  k = 1 + k  k
X=1
Add ab to both sides of the equation :
X ab + ab = a  ab + ab
X=a  ab + ab
X = a
Subtract c from both sides of the equation :
X + c  c = c  b  c
X =  b
Add 2a to both sides of the equation :
X  2a + 2a = a  ab + 2a
a + 2a add up to 3a
X = 3a  ab or
X = a(3  b)
Subtract cb from both sides of the equation :
X + cb  cb = 3cb  c  cb
X = 3cb  c  cb
X = 2cb  c
X = c(2b 1)
Add 5 to both sides of the equation :
X  5 + 5 + a = 2a  5 + 5
X + a = 2a
Subtract a from both sides of the equation :
X + a  a = 2a  a
X = a
Subtract 3 from both sides of the equation :
X + 3  3  k = 6 3k  3
X  k = 3  3k
Add k to both sides of the equation :
X  k + k = 3  3k + k
X = 3  2k
Add 1 to both sides of the equation :
X  1 + 1  ab = ab  1 + 1;
X  ab = ab
Add ab to both sides of the equation :
X  ab + ab = ab + ab
X = 2 ab
Add a to both sides of the equation :
Xa + a  b=ab + a
X  b = 2a  b
Add b to both sides of the equation :
X  b + b = 2a  b + b
X = 2a
Subtract 2a from both sides of the equation :
X + 2a  2a  3c = 3a  2c  2a.
X  3c = a  2c
Add 3c to both sides of the equation :
X  3c + 3c = a  2c + 3c
X = a + c
APPENDIX 2.
PRACTICE 2. SOLUTIONS.
Add 3 to both sides of the equation :
2X  3 + 3 = 5 + 3
2X = 8
Divide both sides of the equation by 2:
2X/2 = 8/2
X=4
Add 5 to both sides of the equation :
3X  5 + 5 = 4 + 5
3X = 9
Divide both sides of the equation by 3:
3X/3 = 9/3
X=3
Subtract 6 from both sides of the equation :
5X + 6  6 = 36  6
5X = 30
Divide both sides of the equation by 5:
5X/5 = 30/5
X=6
Add 5 to both sides of the equation :
8X  5 + 5 = 43 + 5
8X = 48
Divide both sides of the equation by 8:
8X/8 = 48/8
X=6
Add 2 to both sides of the equation :
7X  2 + 2 = 19 + 2
7X = 21
Divide both sides of the equation by 7:
7X/7 = 21/7
X=3
Subtract 8 from both sides of the equation :
4X + 8  8 = 20  8
4X = 12
Divide both sides of the equation by 4:
4X/4 = 12/4
X=3
Add a to both sides of the equation :
6X  a + a = 2a + a
6X = 3a
Divide both sides of the equation by 6:
6X/6 = 3a/6
X = a / 2
Subtract b from both sides of the equation :
2X + b  b = 13b  b
2X = 12b
Divide both sides of the equation by 2:
2X/2 = 12b/2
X = 6b
Subtract 3a from both sides of the equation :
7X + 3a  3a = a + b  3a
7X = b  2a
Divide both sides of the equation by 7:
7X/7 = (b  2a)/7
X = (b  2a)/ 7
Add 2a to both sides of the equation :
4X  2a + 2a = 4 + 2a + 2a
4X = 4 + 4a
Divide both sides of the equation by 4:
4X/4 = 4/4 + 4a/4
X = 1 + a
Add 3a to both sides of the equation :
4X  3a + 3a = a + 3a
4X = 4a
Divide both sides of the equation by 4:
4X/4 = 4a/4
X = a
Add 2b to both sides of the equation :
3X  2b + 2b = 6  14b + 2b
3X = 6  12b
Divide both sides of the equation by 3:
3X/3 = (6  12b)/3
3X/3 = 6/3  12b/3
X = 2  4b
Add 2a to both sides of the equation :
6X  2a + 2a = 24b  20a + 2a
6X = 24b  18a
Divide both sides of the equation by 6:
6X/6 = (24b  18a)/6
6X/6 = 24b/6  18a/6
X = 4b  3a
Add 3a to both sides of the equation :
aX  3a + 3a = ab  2a + 3a
aX = ab + a
Factor a out of the term: ab + a
aX = a(b + 1)
Divide both sides of the equation by a:
aX/a = a(b + 1)/a
a should not be equal to 0!
X = b + 1
Subtract ab from both sides of the equation :
2aX + ab  ab = 2a  ab  ab
2aX = 2a  2ab
2aX = 2a(1  b)
Divide both sides of the equation by 2a:
2aX/2a=2a(1  b)/2a
X=1b
Add c to both sides of the equation :
3aX  c + c = 3ac  7c + c
3aX = 3ac  6c
Divide both sides of the equation by 3a:
3aX/3a = (3ac  6c)/3a
3aX/3a = 3ac/3a  6c/3a
X = c  2c/a
APPENDIX 2.
PRACTICE 3. SOLUTIONS.
Subtract 1 from both sides of the equation :
1  X  1 = 5  a  1
X = 4  a
Multiply both sides of the equation by (1):
(X)(1) = (1) (4  a)
X = ( 4 + a)
X = a  4
Subtract 1 from both sides of the equation :
1  2X  1 = X  4  1
 2X = X  5
Subtract X from both sides of the equation :
 2X  X = X  5  X
 3X =  5
Divide both sides of the equation by 3:
 3X/(3) =  5/(3)
X = 5/3
Subtract a from both sides of the equation :
a  3X  a = b  X  a
3X = b  X  a
Add X to both sides of the equation :
3X + X = b  X  a + X
2X = b  a
Multiply both sides of the equation by (1):
2X (1) = (1) (b  a)
2X = a  b
Divide both sides of the equation by 2:
2X/2 = (a  b)/2
X = (a  b) / 2
Subtract 2a from both sides of the equation :
2a  4X  2a = 2X  4a  2a
 4X = 2X  6a
Subtract 2X from both sides of the equation :
4X  2X = 2X  2X  6a
6X =  6a
Multiply both sides of the equation by (1):
(6X)(1) = ( 6a)(1)
6X = 6a
Divide both sides of the equation by 6:
6X/6 = 6a/6
X = a
Add 2X to both sides of the equation :
4b  2X + 2X = 2X  4b + 2X
4b = 4X  4b
Add 4b to both sides of the equation :
4b + 4b = 4X  4b + 4b
8b = 4X
Divide both sides of the equation by 4:
8b/4 = 4X/4
2b = X
X = 2b
Subtract aX from both sides of the equation :
ab + aX  aX = 2aX + ac  aX
ab = 2aX  aX + ac
ab = aX + ac
Subtract ac from both sides of the equation :
ab  ac = aX + ac  ac
ab  ac = aX
Factor a out of the term: ab  ac
a(b  c)= aX
aX = a(b  c)
Divide both sides of the equation by a:
aX/a = a(b  c)/a
Where a is not equal to 0!
X = a (b  c)/a
X = b  c
Subtract ab from both sides of the equation :
ab + aX  ab=2aX ac  ab
aX = 2aX ac  ab
Subtract aX from both sides of the equation :
aX  aX = 2aX ac  ab  aX
0 = 2aX ac  ab  aX
0= aX  ac  ab
Add ac + ab to both sides of the equation :
0 + ac + ab = aX  ac  ab + ac + ab
ac + ab = aX
Factor a out of the term: ac + ab
a(c + b) = aX
Divide both sides of the equation by a:
a(b + c)/a = aX/a
Where a is not equal to 0!
b + c = X
X = b + c
APPENDIX 2.
PRACTICE 4. SOLUTIONS.
Add 2ab to both sides of the equation :
bX  2b + 2b = aX  2a + 2b
bX = aX  2a + 2b
Subtract aX from both sides of the equation :
bX  aX = aX  2a + 2b  aX
bX  aX = 2b  2a
Factor X out of term: bX  aX
X(ba) = 2(b  a)
Divide both sides of the equation by (b  a):
X(ba)/(b  a) = 2(b  a)/(b  a)
Where (ba) does not equal to 0!
X = 2(ba)/(ba) = 2
Subtract b from both sides of the equation :
b  b  2bX = a  b  2aX
2bX = a  b  2aX
Add 2aX to both sides of the equation :
2bX + 2aX = a  b  2aX + 2aX
2bX + 2aX = a  b
2aX  2bX = a  b
2X(a  b) = (a  b)
Divide both sides of the equation by (a  b):
2X(a  b)/(a  b) = (a  b)/(a  b)
2X = 1
Divide both sides of the equation by 2:
2X/2 = 1/2
X = 0.5
Factor X out of term: aX  bX
X(a  b) = 1
Divide both sides of the equation by (a  b):
X(a  b)/(a  b) = 1/(a  b)
X = 1/(a  b)
4. aX  bX  cX = 2a  2b  2c
X(a  b  c) = 2(a  b  c)
Divide both sides of the equation by (a  b  c):
X(a  b  c)/(a  b  c)=2(a  b  c)/(a  b  c)
X = 2
Add 5a to both sides of the equation :
3abX  5a + 5a = 3acX + 13a + 5a
3abX = 3acX + 18a
Subtract 3acX from both sides of the equation :
3abX  3acX = 3acX + 18a  3acX
3abX  3acX = 18a
Factor 3aX out of term: 3abX  3acX
3aX(b  c) = 18a
Divide both sides of the equation by (b  c):
3aX(b  c)/(b  c) = 18a/(b  c)
3aX = 18a/(b  c)
Divide both sides of the equation by 3a:
3aX/3a = 18a/3a (b  c)
X = 6/(b  c)
Factor X out of term: aX  bX
X(a  b) = c(a  b)
Divide both sides of the equation by (a  b):
X(a  b)/(a  b) = c(a  b)/(a  b)
X = c
Add 4X to both sides of the equation :
9a  4X + 4X = 5a  2X + 4X
9a = 5a  2X + 4X
9a = 5a + 2X
Subtract 5a from both sides of the equation :
9a  5a = 5a + 2X  5a
4a = 2X
2X = 4a
Divide both sides of the equation by 2:
2X/2 = 4a/2
X = 2a
Factor X out of term: X  aX
X (1  a) = 2(1  a)
Divide both sides of the equation by (1  a):
X (1  a)/(1  a) = 2(1  a)/(1  a)
X= 2(1  a)/(1  a)
X = 2
Factor an "X" out of term:aX  bX
X (a  b) = b  a
Factor out (1) from the right side of the equation.
X (a  b) = (b + a)(1)
X (a  b) = (a  b)(1)
Divide both sides of the equation by (a  b):
X(a  b)/(a  b) = (1)(a  b)/(a  b)
X = 1(a  b)/(a  b)
X = 1
APPENDIX 2.
PRACTICE 5. SOLUTIONS.
Factor 5X out of term: 5aX  5bX
5X (a  b) = 10b  10a
Factor 10 out of term: 10b  10a
5X (a  b) = 10(b + a)
Divide both sides of the equation by (a  b):
5X(a  b)/(a  b) = 10(a  b)/(a  b)
5X = 10;
5X/5 = 10/5
X =  2
Factor X out of term: aX  bX  cX
X (a  b  c) = c + b  a
X (a  b  c) = ( c  b + a) (1)
X(a  b  c) = (a  b  c)(1)
Divide both sides of the equation by (a  b  c):
X(a  b  c)/(a  b  c) = (a  b  c)(1)/(a  b  c)
X = 1
Factor X out of term: 2X  3aX
X (2  3a) = 2(3a  2)
X (2  3a) =  2( 3a + 2)
X (2  3a) = 2 (2  3a)
Divide both sides of the equation by (2  3a):
X (2  3a)/(2  3a) = 2 (2  3a)/(2  3a)
X =  2
Factor X out of term: 3aX  9bX
3X (a  3b) = 9(3b  a)
3X (a  3b) = 9(a  3b)
Divide both sides of the equation by (a  3b):
3X (a  3b)/(a  3b) = 9(a  3b)/(a  3b)
3X = 9
Divide both sides of the equation by 3:
3X/3 = 9/3
X =  3
Factor X out of term: 4bX  cX
Factor 8 out of term: 8c  32b
X (4b  c) = 8(c  4b)
X (4b  c) = 8(c + 4b)
Divide both sides of the equation by (4b  c):
X (4b  c)/(4b  c) = 8(c + 4b)/(4b  c)
X =  8
Factor X out of term: abX  acX
Factor a out of the term: ac  ab
aX (b  c) = a(c  b)
aX (b  c) = a(c + b)
aX (b  c) = a ( b  c)
Divide both sides of the equation by (b  c):
aX (b  c)/(b  c) = a ( b  c)/(b  c)
Where (bc) does not equal to 0!
aX = a
Divide both sides of the equation by a:
aX/a = a/a
Where a is not equal to 0!
X =  1
Multiply both sides of the equation by 2:
2(X/2  aX) = 2 (1  2a)
X  2aX = 2 (1  2a)
Factor X out of term: X  2aX
X(1  2a)= 2(1  2a)
Divide both sides of the equation by (1  2a):
X(1  2a)/(1 2a)= 2(1  2a)/(1 2a)
X = 2
Subtract 2a from both sides of the equation :
aX/5 + 2a  2a = 5a  4aX  2a
aX/5 = 5a  4aX  2a
Add 4aX to both sides of the equation :
aX/5 + 4aX = 5a  4aX  2a + 4aX
aX/5 + 4aX = 5a  2a
aX/5 + 4aX = 3a
Multiply both sides of the equation by 5:
5(aX/5 + 4aX) =5 * 3a
aX + 20aX = 15a
21aX = 15a
Divide both sides of the equation by a:
21aX/a = 15a/a
Where a is not equal to 0!
21X = 15
Divide both sides of the equation by 21:
21X/21 = 15/21
X = 15/21
X = 5/7
APPENDIX 2.
PRACTICE 6. SOLUTIONS.
X  Y =  5
We have two equations. Write one equation on top of the other and find their sum.
if 2X =  4 then 2X / 2 =  4 / 2 and
X =  2
To find Y, substitute X with (2) in one of equations. X + Y = 1
(2) + Y = 1
Then (2) + 2 + Y = 1 + 2
Y = 3
2X  2Y = 6
In the first equation we have positive Y and in the second equation we have negative 2Y.
To get rid of Y in an equation we can multiply the first equation by 2.
We will get 2X + 2Y = 2.
Now we can add the equations
4X = 8 then X =2
Substitute X with 2 in the first equation:
2+Y=1 then Y= 12 then Y = 1.
2X + Y =  4
In the first equation we have X and in the second 2X.
To get rid of X we can multiply the first equation by 2
2(X + 2Y) =2 *1 then we get 2X + 4Y = 2.
Now we subtract the second equation from the first equations.
We get 6 because 2  ( 4 ) = 6
Then 3Y = 6 and Y=2. Substitute Y with 2 in the first equation
X + 4=1 then X =  3
Substitute X and Y with their values in the first equation:
( 3) + 2 * 2 = 1 then  3 + 4 = 1 and 1=1
X + 3Y = 0
Multiply the second equation by 2.
2(X + 3Y) = 2 * 0 then 2X + 6Y = 0
Subtract the new second equation from the first one.
Then 5Y/5 = 5/5 and Y = 1
Substitute Y in the first equation with 1
Then 2X  1 = 5 then 2X = 6 and X = 3.
Substitute X and Y with their values in the first equation:
2 * 3  1 = 5 5 = 5
4X + 2Y = 10
Multiply the first equation by 2 then we will get
6X  2Y =10.
Add the second equation to the first one
Then X = 20/10 = 2
Substitute X in the first equation:
3*2  Y = 5
6  Y = 5. Subtract 6 from both sides of the equation and you get:
Y = 1
Multiply both sides of the equation with 1 and you get:
Y = 1
Substitute X and Y in the second equation:
4*2 + 2* 1 = 10 and 8 + 2 = 10 10=10
Substitute X and Y in the first equation:
3*2  1 = 5 5 = 5
4X  2Y = 6
Now add the equations:
8X/8 = 16/8 and X = 2
Substitute X in the first equation.
4 * 2 + 2Y = 10 You get:
8 + 2Y = 10
Subtract 8 from both sides of the equation:
8 – 8 + 2Y = 10  8
2Y = 2
Y = 1
Substitute X and Y in the second equation:
4 * 2 – 2* 1 = 6
8 – 2 = 6
6 = 6
Substitute X and Y in the first equation:
4 * 2 + 2 * 1 = 10
8 + 2 = 10
10 = 10
7X + 2Y= 6
Multiply the first equation by 2 and get:
2X  2Y = 3
Sum the equations:
9X=9 then X = 1
Substitute X in the first equation:
1  Y = 1.5 Then Y = 0.5 and Y = 0.5
Substitute X and Y in the first equation:
1  ( 0.5) = 1.5 and 1 + 0.5 = 1.5 1.5 = 1.5
Substitute X and Y in the second equation:
7*1 + 2(0.5) = 7  1 = 6 and 6 = 6
X  3Y = 3
Multiply the first equation by 3 and get:
9X + 3Y = 27
Sum the equations:
X = 30/10 = 3
Substitute X in the first equation and find Y:
3*3 + Y = 9
Y = 99 = 0
Substitute X and Y in the first equation:
3*3 + 0 = 9 and 9=9
Substitute X and Y in the second equation:
3  3*0 = 3 and 3 = 3
X + 3Y = 2
Multiply each part of the second equation by 5 and get:
5X  15Y = 10
Sum the equations:
17Y = 17 then Y = 1
Substitute Y in the second equation and find X:
X + 3*1 = 2
then X  3 + 3 = 2  3
X =  1
Substitute X and Y in the first equation:
5(1)  2*1 = 7 then
5  2 = 7 and 7 = 7
Substitute X and Y in the second equation:
1 + 3*1 = 2 then 1 + 3 = 2 and 2 = 2
X  2Y =  1
Multiply the second equation by 9 and get
9X  18Y = 9
Subtract the second equation from the first one:
We get 21Y because 3Y  (18Y)=21Y
the same way we get 21 (12 (9)=21
Substitute Y in the second equation and find X:
X  2*1 = 1 then
X = 1 + 2 = 1
Substitute X and Y in the first equation:
9*1 + 3*1 = 12 12 = 12
Substitute X and Y in the second equation:
1 2*1 = 1 then 1  2 = 1 and 1 = 1
APPENDIX 2.
PRACTICE 6. GRAPHICAL SOLUTIONS.
To solve simultaneous equations graphically we have to plot a graph for each equation. The intersection point of these two lines gives us the solution.
1. X + Y = 1
X  Y =  5
Find 2 points to draw the first line: X + Y = 1
Modify the equation to a general form Y = aX + b
Subtract X from both sides of the equation.
X  X+ Y = 1  X
Y = 1  X
X= 0 Y(0)= 1  0 = 1
X= 5 Y(5) = 1  5 =  4
Find 2 points to draw the second line: X  Y =  5
Modify the equation to a general form Y = aX + b
Subtract X from both sides of the equation.
X  Y  X =  5  X
 Y =  5  X
Multiply both sides by 1 and you get:
Y = 5 + X
X = 0 Y(0) = 5 + 0 = 5
X = 5 Y(5) = 5  5 = 0
Graph 1. The point of the intersection: X=  2, Y= 3.
2. X + Y = 1
2X  2Y = 6
Y = 1  X
The first line points:
X = 0; Y(0) = 1
X = 5; Y(5) = 1  5 =  4
The second line points:
2Y = 2X + 6
Divide both sides by 2
Y = X  3
X = 0; Y(0)= 3
X = 3; Y(3)= 3  3 =0
Graph 2. The point of the intersection: X= 2, Y= 1.
3. X + 2Y = 1
2X + Y =  4
The first line points: X + 2Y = 1
2Y = 1  X
Y = (1  X)/2
X = 5; Y(5)=(1  5)/2 =  2
X = 5; Y(5)=(1 ( 5)/2 =3
The second line point: 2X + Y =  4
Y =  4  2X
X = 0; Y(0) =  4
X = 4; Y(4)=  4  2 ( 4)=  4 +8 = 4
Graph 3. The point of the intersection: X=3, Y=2.
4. 2X + Y = 5
X + 3Y = 0
The first line points: 2X + Y = 5
2X  2X + Y = 5  2X
Y = 5  2X
X = 0; Y(0) = 5
X = 3; Y(3) = 5  2*3 = 1
The second line points: X + 3Y = 0
X  X + 3Y =  X
3Y =  X
Y =  X / 3
X = 6; Y(6) = 2
X = 3; Y(3) = 1
Graph 4. The point of the intersection: X=3, Y=  1.
5. 3X  Y = 5
4X + 2Y = 10
The first line points: 3X  Y = 5
3X  3X  Y = 5  3X
Y = 5  3X
Y = 3X  5
X = 0; Y(0)= 5
X = 2; Y(2)= 3*2 5 =1
The second line points: 4X + 2Y = 10
Subtract 4X from both sides of the equation.
4X  4X + 2Y= 10  4X
2Y = 10  4X
Divide both sides of the equation by 2.
2Y/2 = 10/2  4X/2
Y = 5  2X
X = 0; Y = 5
X = 4; Y = 5 2*4= 3
Graph 5. The point of the intersection: X=2, Y=1.
6. 4X + 2Y = 10
4X  2Y = 6
The first line points: 4X + 2Y = 10
Subtract 4X from both sides of the equation.
4X  4X + 2Y = 10  4X
2Y = 10  4X
Divide both sides by 2.
2Y/2 =10/2  4X/2
Y = 5  2X
X = 0; Y (0) = 5  2*0 = 5
X = 5; Y (5) = 5  2 * 5=5 – 10 = 5
The second line points: 4X  2Y = 6
4X  4X  2Y = 6  4X
2Y = 6 – 4X
2Y/2 = 6/2  4X/2
 Y = 3 – 2X
(1)(Y) = (1)(3 – 2X)
Y = 2X 3
X = 0; Y (0) = 2*0 3 = 3
X = 4; Y (4) = 2*4 3 = 5
Graph 6. The point of the intersection: X= 2, Y=1.
7. X  Y=1.5
7X + 2Y=6
The first line points: X  Y=1.5
Subtract X from both sides of the equation.
X  X  Y = 1.5  X
 Y = 1.5 – X
Multiply both sides by 1
(1)(Y) = (1) (1.5  X)
Y = (1.5 + X)
Y = X  1.5
X =  2; Y (2) = 2 – 1.5= 3.5
X= 5; Y (5) = 5 1.5= 3.5
The second line points: 7X + 2Y=6
Subtract 7X from both sides of equation.
7X  7X + 2Y= 6  7X
2Y= 6  7X
Divide both sides of the equation by 2
2Y/2= 6/2  7X/2
Y = 3  7X/2
X = 0; Y (0) = 3
X = 2; Y (2) = 3  7*2/2 = 3  7 =  4
Graph 7. The point of the intersection: X=1, Y=0.5.
8. 3X + Y = 9
X  3Y = 3
The first line points: 3X + Y = 9
Subtract 3X from both sides of the equation.
3X  3X + Y = 9  3X
Y = 9  3X
X = 4; Y (4) =9 3*4=9 – 12 = 3
X= 2; Y (2) =9  3*2= 3
The second line points: X  3Y = 3
Subtract X from both sides of the equation.
X  X  3Y = 3  X
 3Y = 3  X multiplied by 1
3Y = X  3
Y = (X 3)/3
X = 0; Y (0) =  1
X =  3; Y (3) = ( 3 – 3)/3 = 2
Graph 8. The point of the intersection: X=3, Y=0.
9. 5X  2Y = 7
X + 3Y= 2
The first line points: 5X  2Y = 7
Subtract 5X from both sides of the equation.
5X  5X  2Y = 7  5X
2Y =  7  5X
Multiply both sides of the equation by 1 and you get:
2Y =7 + 5X
Divide both sides of the equation by 2 and you get:
Y = (7 + 5X)/2
X = 0; Y(0)= 7/2 =3.5
X = 1; Y(1)= (7 + 5)/2 =6
The second line points: X + 3Y= 2
X  X + 3Y = 2  X
3Y = 2  X
Y = (2  X)/3
X =  4; Y( 4)= (2  (4))/3= 2
X= 1; Y(1)=(2 (1)/3=1
X=8; y(8)= (2  8)/3=2
Graph 9. The point of the intersection: X= 1, Y=1.
10. 9X + 3Y = 12
X  2Y =  1
The first line points: 9X + 3Y = 12
9X  9X + 3Y = 12
3Y = 12  9X
Divide both sides of the equation by 3 and you get:
Y = 4  3X
X = 0; Y(0) = 4
X = 1; y(1)= 4 3(1)=7
The second line points: X  2Y =  1
Subtract X from both sides of the equation.
X  X  2Y = 1  X
2Y =  1  X
Divide both sides of the equation by 2.
Y = (1 + X)/2
X = 0; Y(0)= 0.5
X = 3; Y(3) = (1 + 3)/2=2
Graph 10. The point of the intersection: X=1, Y=1.
APPENDIX 2.
PRACTICE 7. SOLUTIONS.
Divide each part of the equation by 3 and get:
X^2  25 = 0
Factor the resulting equation:
(X  5)(X + 5) = 0
then X  5= 0 or X = 5
X + 5 = 0 and X = 5.
Answer is X may be 5 or 5.
Let us check if X^2  25 = 0
(5)(5) = 25 and 5 * 5 = 25 We are correct.
Factor the equation and get:
(2X  1) (X  4) =0 To check
multiply (2X  1) (X  4)
and you will get the same equation:
2X^2  8X  1X +4 =
=2X^2 9X +4
Then 2X  1 = 0 and X = 0.5
Find the second X value:
X  4 = 0
X = 4
Check X=0.5:
2 * (0.5)(0.5)  9 (0.5) + 4 =0
2* 0.25  4.5 + 4 =0
and 0.5  4.5 + 4 =0
4.5  4.5 =0.
Check X = 4
2 * 4 * 4  9 * 4 + 4 = 0
then 32  36 + 4 = 0 and  4 + 4 =0
Factor the equation:
(3X + 1)(X 2) =0
Check 3X * X  6X + X  2 =
=3X^2 5X 2=0 We are correct.
3X + 1 = 0 then 3X = 1 and X = 1/3
X  2 = 0 then X = 2
Let us substitute the first X in the equation:
3 * (1/3)(1/3)  5 (1/3) 2 =
=3/9 + 5/3 2 =
=3/9 + 15/9 18/9 = 18/9  18/9 =0
Let us substitute the second X in the equation:
3*2*2  5*2  2 =0 then 12 10 2=0
4. 4X^2 13X +3=0 Factor the equation
(4X  1)(X  3) = 0
To check multiply: (4X  1)(X  3)
and you will get the same equation:
4X*X  12X  X + 3 =
=4X^2 13X +3
4X 1 = 0 then X = 0.25
X  3 = 0 then X = 3
Let us substitute the first X in the equation:
4*0.25*0.25  13*0.25 +3 =
=0.25  3.25 + 3= 0
We are correct.
Let us substitute the second X in the equation:
4*3*3 13*3 +3 =
=36  39 +3 =0
We are correct
5. 7X^2  29X +4=0 Factor the equation:
(7X  1)(X  4) =0
Check: (7X  1)(X  4)=
=7X*X  28X  X + 4 =
7X^2  29X + 4
7X 1 = 0 then
X = 1/7; X4 = 0
then X = 4
Let us substitute the first X in the equation:
7 (1/7)(1/7)  29(1/7) + 4 =0
7/49  29/7 + 4 =
=7/49  29*7/49 + 4*49/49=
7/49  203/49 + 196/49=
203/49  203/49 =0
Let us substitute the second X in the equation
7*4*4  29*4 + 4 =
=112  116 + 4=0
We are correct
6. 5X^2 28X +15 Factor the equation:
(5X  3)(X  5) =0
Check: (5X  3)(X  5) =
=5X*X  25X  3X + 5 =
5X^2 28X +15
5X 3 =0 then X =3/5
X  5 = 0 then
X = 5
Let us substitute the first X in the equation:
5*(3/5)(3/5)  28(3/5) + 15 =
=5 (9/25)  28 (3/5) + 15 =
=9/5  84/5 + 75/5 =
75/5 + 75/5 =0
We are correct.
Let us substitute the second X in the equation:
5* 5*5 28*5 +15 =
=125  140 + 15=0
We are correct.
7. 18X^2 +12X 6=0 Factor the equation
(3X + 3)(6X  2)
Check: (3X + 3)(6X  2) =
18X*X  6X + 18X  6 =
18X*X + 12X  6 =
18X^2 +12X 6
3X + 3 =0 then
X = 1
6X 2 = 0
then X = 1/3
Let us substitute the first X in the equation:
18 * (1)(1) + 12(1)  6 =
=18  12  6 =0
Let us substitute the second X in the equation:
18 (1/3)(1/3) + 12(1/3)  6 =
=18/9 + 12/3  6 =
6/3 + 12/3  18/3 =
18/3  18/3 =0
We are correct.
8. 3X^2 12X + 9=0 Factor the equation
(3X  3)(X 3)=0
Check: (3X  3)(X 3) =
=3X * X  9X  3X + 9 =
3X * X  12X + 9 =
3X3 = 0 then X =1
X  3 = 0 then X=3
Let us substitute the first X in the equation:
3* 1*1 12*1 +9 =
=3  12 + 9 =0
We are correct.
Let us substitute the second X in the equation:
3*3*3 12*3 + 9 =
=27  36 +9 = 0
We are correct.
9.24X^2 +55X 24=0 Factor the equation:
(8X  3)(3X + 8)
Check: (8X  3)(3X + 8)=
=24X * X + 64X  9X  24 =
=24X * X + 55X  24 =
=24X^2 + 55X 24
8X 3 = 0 then
X = 3/8
3X + 8 = 0
then X = 8/3
Let us substitute the first X in the equation:
24* 3/8*3/8 + 55*3/8 24=
= 24* 9/64 + 165/8 24 =
3*9/8 + 165/8  24*8/8=
27/8 + 165/8 192/8 =
192/8  192/8 = 0 We are correct.
Let us substitute the second X in the equation:
24 * (8/3)(8/3) + 55(8/3)  24=
=24 (64/9)  440/3  24 =
512/3  440/3  24* 3/3 =
=512/3  440/3  72/3 = 0
We are correct.
10.12X^2 45X 12=0 Factor the equation:
(4X + 1)(3X  12) =0
Why did not I choose
(4X + 3)(3X  4) ? 4X * 3X =
=12X^2 and 3 * (4) = 12.
I did not choose (4X + 3)(3X  4) and have chosen
(4X + 1)(3X  12) instead because in the equation we have
45X as a second member.
(4X + 3)(3X  4) can give us 16X (4X * 4) and 9X (3*3X)
We cannot get 45X from (4X + 3)(3X  4)
I have chosen (4X + 1)(3X  12)
because 4X * 12 gives us 48X and it is close to 45X.
Let us check:
(4X + 1)(3X  12) =
=12X*X  48X + 3X  12 =
12X*X  45X 12 =
=12X^2 45X 12
4X + 1 = 0
Then X =  0.25
3X  12 = 0
Then X = 12/3=4
Let us substitute the first X in the equation:
12*(0.25)(0.25)  45(0.25)  12 =
0.75  45*(0.25)  12 =
=0.75 + 11.25  12 =
=12  12 =0
We are correct.
Let us substitute the second X in the equation:
12* 4*4  45*4 12 =
=192  180  12=0
We are correct.
11.8X^2 28X +12=0 Factor the equation:
(4X  2)(2X  6)
Check: (4X  2)(2X  6) =
4X*2X  24X  4X + 12 =
=8X^2 28X +12
4X  2 = 0 then X = 0.5
2X6 =0 then X =3
Let us substitute the first X in the equation:
8*(0.5)(0.5)  28*(0.5) + 12=
=2  14 + 12 =0
We are correct
Let us substitute the second X in the equation:
8*3*3 28*3 +12 =
=72  84 + 12=0
We are correct.
Divide the equation by 12 and get
X^2  4=0
Factor the equation: (X  4) (X + 0) =0
Check: (X  4) (X + 0) =
=X*X + 0 + 4X  0) =
=X^2  4=0
X  4 = 0 then X = 4
X + 0 = 0 then X =0
Let us substitute the first X in the equation:
12*4*4  48*4 =0 we are correct.
Let us substitute the second X in the equation:
12*0  48*0 = 0 we are correct.
13.2X^2  11X + 9=0 Factor the equation:
(2X  9)(X  1)
Check: (2X  9)(X  1) =
=2X*X  2X  9X + 9 =
2X^2  11X + 9
(2X  9) = 0 then X =4.5
X  1 = 0 then X =1
Let us substitute the first X in the equation:
2* 4.5 * 4.5  11*4.5 +9 =
= 40.5  49.5 + 9=0
We are correct.
Let us substitute the second X in the equation:
2 * 1*1  11*1 +9 =
= 2  11 + 9 =0
We are correct.
14.14X^2  27X + 9 = 0 Factor the equation:
(2X  3)(7X  3)
Check: (2X  3)(7X  3) =
2X*7X  21X  6X + 9 =
=14X*X  27X +9 =
=14X^2  27X +9
2X  3 = 0 then X = 1.5
7X 3 = 0 then X = 3/7
Let us substitute the first X in the equation:
14*1.5*1.5  27*1.5 + 9 =
31.5  40.5 + 9 = 0
We are correct Let us substitute the second X in the equation:
14*3/7*3/7  27*3/7 +9=
=14*9/49  27*3/7 +9=
=2*9/7  27*3/7 +9=
18/7  81/7 + 63/7 =
81/7  81/7 =0
We are correct.
15.5X^2  37X + 14=0 Factor the equation:
(5X  2)(X  7)
Check: (5X  2)(X  7) =
=5*X*X  35X  2X + 14 =
=5*X*X  37X + 14 =
=5X^2  37X +14
5X  2 = 0 then X = 0.4
X  7 = 0 then X =7
Let us substitute the first X in the equation:
5*0.4*0.4  37*0.4 + 14 =
=0.8  14.8 + 14=0
We are correct.
Let us substitute the second X in the equation:
5 *7*7  37*7 +14 =
=245 259 +14 =0
We are correct.
If you find any errors in an equation or typos, please send an email to tutor@hardstuffez.com
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