ANSWERS
INTERMEDIATE ALGEBRA
TEST CHAPTER 4
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Any problem with a graph to be answered will be found in the math lab.
1.
Substitute (-5) in the place of x and 3 in the place of y in both equations.
2(-5) – 3(3) = 21
4(-5) + 2(3) = 2
-10 –9 =21
-20 + 6 =2
-19 ≠ 21
-14 ≠ 2
(-5, 3) is not a solution to this system.
2.
You will find the answer to this problem in the math lab.
3.
Step a.
Step b.
Step c.
Step d.
Step e.
Step f.
Step g.
Substitute the expression 3x – 13 into the equation x + 2y = 9 in the place of y.
x + 2(3x – 13) = 9
Distribute 2 through the parentheses. x + 6x – 26 = 9.
Collect like terms on the left side of the equation. 7x – 26 = 9.
Add 26 to both sides of the equation. 7x – 26 + 26 = 9 + 26
Collect like terms on the right side of the equation. 7x = 35
Divide both sides of the equation by 7.
7x = 35
7
7
Step h. x = 5
Step i. Substitute 5 in the place of x. I am using the equation 3x – 13 = y.
3(5) – 13 = y
15 – 13 = y
y=2
The answer to this problem is (5,2)
4.
To eliminate the x's multiplying by (-3) in the first term and multiplying by 2 in the second term
OR to eliminate the y's multiply by 2 in the first term and multiply by 5 in the second term.
-3(2x – 5y = (-11))
2(3x + 2y = 12 )
-6x + 15y = 33
6x + 4y = 24
19y = 57
y=3
Substitute 3 in the place of y in either
the first equation or the second equation
ALSO
2(2x – 5y = (-11))
5(3x + 2y = 12)
4x – 10y = (-22)
15x + 10y = 60
19x = 38
x=2
Substitute 2 in the place of x in either
in the first equation or the second equation.
1st equation 2x –5(3) = (-11)
1st equation: 2(2) –5y = (-11)
2x –15 = -11
4 - 5y = -11
2x – 15 + 15 = -11 + 15
4 – 4 – 5y = -11 -4
2x = 4
-5y = -15
2
2
-5
-5
x=2
y=3
YOU ONLY NEED TO DO ONE SIDE OF THE "ALSO". EITHER THE LEFT SIDE OR THE
RIGHT SIDE. DO NOT WORK BOTH SIDES TO FIND THE ANSWERS.
Answer to this problem is (2,3)
page 2
In problems 5 – 7, the first step is to get all equations into the slope y-intercept formula. y = mx + b
5.
Step a. Multiply the first equation by 15 to eliminate the fractions.
15( 1/5x - ⅓y = 0)
3x –5y =0
Step b. Use the addition method to eliminate the y's
3x – 5y = 0
5y = 3x
3x = 3x
3x – 3x = 3x – 3x
0x = 0
When the coefficient is zero and the result of the answer is zero, then the 2 equations are the SAME LINE
AND THEY COINCIDE.
6.
Step a. Distribute (-4) through the parenthesis.
3x –4y + 4 =0
3x –4y + 4y + 4 = 0 + 4y
3x + 4 = 4y
4 4 4
¾x + 1 = y
Step b. Rewrite in the form of y = mx + b
5x + 14 = 3y
3
3 3
5/3x + 14/3 = y
Step c. The slope in the first equation is ¾ and the slope in the second equation is 5/3. Since the
slopes are different, THE LINES WILL INTEREST.
7.
Step a. Multiply the 1st equation by 6, to eliminate the denominators.
6(½x + ⅓y = ⅔)
3x + 2y = 4
Step b. Rewrite the equation in the form of y = mx + b
3x – 3x + 2y = 4 – 3x
2y = 4 – 3x
2
2 2
y = 2 – 3/2x
Step c. Multiply the 2nd equation by 15, to eliminate the denominators.
15(⅓x + 2/5y = 4/15)
5x + 6y = 4
Step d. Rewrite in the form of y= mx + b
5x – 5x + 6y = 4 – 5x
6y = 4 – 5x
6 6 6
y=
2 5
 x
3 6
 5
 , THEREFORE THE
 6 
The slope for the 1st equation is (-3/2) and the slope for the 2nd equation is 
SLOPES ARE DIFFERENT AND THE LINES INTERSECT.
Problems 8- 10, you will find the answers in the math lab.