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# solutions-Derivita SAT PREP Diagnostic test Linear Equations

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```2/3/22, 12:42 PM
Derivita: SAT PREP Diagnostic test: Linear Equations
SAT PREP Diagnostic test: Linear Equations
Name
Date
Student ID
1.
Section
Suppose a well-designed survey shows that orange​is the favorite color for 4​out of every 9​
students at a particular school. If there are 585​students at the school, what would be the
expected number of students at the school whose favorite color is orange​?
students
Solution
We will begin by letting x =​the number of students at the school whose favorite color is orange​. Then, we
will translate the problem into a proportion, taking care that the units are in the same location o​n each
ratio.
4 like orange
x like orange
=
9 students
585 students
​
​Next, we will multiply both sides of the equation by the LCD, 585​students and solve for x​as follows.
585 students ⋅
4 like orange
9 students
= 585 students ⋅
​
x like orange
585 students
65 (4 like orange) = 1 (x like orange)
​
​
​
260 = x
​Therefore, we could expect approximately 260​students to say their favorite color is orange​.
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Derivita: SAT PREP Diagnostic test: Linear Equations
SAT PREP Diagnostic test: Linear Equations
2.
Solve for n​.
7
13
=
2
4
n−
​
​
n=
Solution
To get n​by itself, we use the Addition Property of Equality. Add
7
​to each side to ‘undo’ the
2
​
subtraction, then simplify.
7
2
7
+
2
n −
​
​
​
13
4
7
+
2
=
​
​
​
​
​
27
4
n =
We can check our work by plugging
​
​
​
27
​in for n​in the original equation:
4
​
27
7
−
4
2
​
27
14
−
4
4
​
13
4
​Since we have a true statement, we know n =
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​
?
=
​
?
​
​
=
✓
​
=
13
4
13
4
13
4
​
​
​
​
27
​is a valid solution.
4
​
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Derivita: SAT PREP Diagnostic test: Linear Equations
SAT PREP Diagnostic test: Linear Equations
3.
Solve for x​.
x + 4.9 = −8.25
x=
Solution
To get x​by itself, we use the Subtraction Property of Equality. Subtract 4.9​from each side to ‘undo’ the
addition, then simplify.
x +4.9 =
−8.25
−4.9
−4.9
x = −13.15
We can check our work by plugging −13.15​in for x​in the original equation:
?
−13.15 + 4.9 = −8.25
✓
−8.25 = −8.25
​Since we have a true statement, we know x = −13.15​is a valid solution.
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Derivita: SAT PREP Diagnostic test: Linear Equations
SAT PREP Diagnostic test: Linear Equations
4.
Use the phrase below to answer the following questions.
Six​less​than m​is equal to 28​.
Translate the phrase into an algebraic equation.
Equation
Now, use your equation to solve for m​.
m=
Solution
The phrase “less than” tells us that we are subtracting 6​from m​, and the phrase “is equal to” indicates an
‘=​’ sign. Therefore, the algebraic equation is
m − 6 = 28
​To solve for m​, add 6​to both sides.
m −6 =
​
​
+6
​
​
m =
28
​
+6
​
​
34
​Be sure to plug this value into the original equation to check that it is a valid solution.
​
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Derivita: SAT PREP Diagnostic test: Linear Equations
SAT PREP Diagnostic test: Linear Equations
5.
Solve for n​.
−n = 26
n=
Solution
Recall that −n​is equivalent to −1n​. To isolate n on one side of the equation we use the Division Property
of Equality. Divide both sides of the equation by −1​.
−n = 26
−1n = 26
−1n
−1
​
​
=
26
−1
​
​
n = −26
​Check this answer by plugging it back in to the original equation.
?
−(−26) = 26
​
✓
26 = 26
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​
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Derivita: SAT PREP Diagnostic test: Linear Equations
SAT PREP Diagnostic test: Linear Equations
6.
Solve for x​.
6
x = 18
7
​
x=
Solution
To isolate x​on one side of the equation we use the Multiplication Property of Equality. Multiply both
6
sides of the equation by the reciprocal of ​.
7
​
6
x = 18
7
​
7 6
7
⋅ x =
(18)
6 7
6
​
​
​
​
​
7 18
1x =
⋅
6
1
​
​
x = 21
​Check this answer by plugging it back in to the original equation.
6
?
(21) = 18
7
​
?
6(3) = 18
✓
18 = 18
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2/3/22, 12:42 PM
Derivita: SAT PREP Diagnostic test: Linear Equations
SAT PREP Diagnostic test: Linear Equations
7.
Solve for y ​.
7 + 12y + 6 − 11y = 25
y=
Solution
To solve for y ​, we first need to simplify each side. Start by rearranging the terms and then combining like
terms.
7 + 12y + 6 − 11y = 25
12y − 11y + 7 + 6 = 25
​
​
y + 13 = 25
​Now that both sides are fully simplified, we can isolate y ​by subtracting 13​from both sides.
y + 13
−13
=
−13
​
y
25
​
=
​
​
12
​We can check our work by plugging 12​in for y ​in the original equation:
?
7 + 12 (12) + 6 − 11 (12) = 25
?
7 + 144 + 6 − 132 = 25
​
​
✓
25 = 25
​Since we have a true statement, we know y = 12​is a valid solution.
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Derivita: SAT PREP Diagnostic test: Linear Equations
SAT PREP Diagnostic test: Linear Equations
8.
Solve for x​.
−3 (5x + 4) + 16 (x − 3) = −2x + 2 (x − 3)
x=
Solution
To solve for x​, we first need to simplify each side. Start by distributing, then rearrange the terms and
combine like terms.
−3 (5x + 4) + 16 (x − 3) = −2x + 2 (x − 3)
−15x − 12 + 16x − 48 = −2x + 2x − 6
​
​
−15x + 16x − 12 − 48 = −6
x − 60 = −6
​Now that both sides are fully simplified, we can isolate x​by adding 60​to both sides.
x − 60
+60
=
​
+60
​
​
=
x
−6
​
​
54
​We can check our work by plugging 54​in for x​in the original equation:
?
−3 (5 (54) + 4) + 16 ((54) − 3) = −2 (54) + 2 ((54) − 3)
?
−3 (274) + 16 (51) = −108 + 2 (51)
​
​
?
−822 + 816 = −108 + 102
✓
−6 = −6
​Since we have a true statement, we know x = 54​is a valid solution.
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Derivita: SAT PREP Diagnostic test: Linear Equations
SAT PREP Diagnostic test: Linear Equations
9.
Solve for n​.
−4 (n − 6) − 3 = 3
n=
Solution
Begin by simplifying each side of the equation, using the distributive property first.
−4 (n − 6) − 3 = 3
−4n − 4 (−6) − 3 = 3
−4n + 21 = 3
​Continue to isolate n​on one side of the equation.
−4n + 21
−21
=
3
−21
​
−4n
=
−18
−4n
−4
=
−18
−4
=
9
2
​
​
n
​
​
​
​
​Be sure to check this answer by plugging it back in to the original equation.
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9/16
2/3/22, 12:42 PM
Derivita: SAT PREP Diagnostic test: Linear Equations
SAT PREP Diagnostic test: Linear Equations
10.
Solve for x​.
−9x = −x + 16
x=
Solution
In this equation, we have variables on both sides. We need to make one side the variable side and the other
the constant side. Since there is already a constant on the right, let's make the variable side be on the left.
To achieve this, we need to add x​to both sides of the equation. Doing so, we get
−9x = −x + 16
−9x + x = −x + x + 16
​
​
−8x = 16
Dividing both sides by −8​and simplifying, we get​
−8x = 16
−8x
−8
​
=
16
−8
​
x = −2
​We can check this solution by plugging it into the original equation.
?
−9 (−2) = − (−2) + 16
✓
18 = 18
​Therefore, x = −2​is a valid solution.
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10/16
2/3/22, 12:42 PM
Derivita: SAT PREP Diagnostic test: Linear Equations
SAT PREP Diagnostic test: Linear Equations
11.
Solve for y ​.
8y + 8 = 12y − 6
y=
Solution
This equation has variables and constants on each side. Since 12 &gt; 8​, let's make the variable side be on the
right and the constant side be on the left. Doing so, we get
8y + 8 = 12y − 6
8y − 8y + 8 = 12y − 8y − 6
8 = 4y − 6
​
​
8 + 6 = 4y − 6 + 6
14 = 4y
Dividing both sides by 4​and simplifying, we get​
14 = 4y
14
4
7
2
​
​
=
4y
4
​
​
=y
​
​We can check this solution by sunstituting it into the original equation.
8⋅
7
7
?
+ 8 = 12 ⋅
−6
2
2
​
​
?
28 + 8 = 42 − 6
​
​
✓
36 = 36
​Therefore, y =
7
​is a valid solution.
2
​
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Derivita: SAT PREP Diagnostic test: Linear Equations
SAT PREP Diagnostic test: Linear Equations
12.
Solve the following equation. If there are multiple answers, enter each answer separated by a
comma. Enter None​if there are no solutions.
∣13x + 5∣ = 6
x=
Solution
This equation will be true any time the expression inside the absolute value bars is equal to 6​or −6​. This
leads us to the following two equations, 13x + 5 = 6​or 13x + 5 = −6​.
Solving the first equation, we get,
13x + 5 = 6
13x + 5 − 5 = 6 − 5
13x = 1
​
13x
13
​
1
=
13
​
1
13
x =
​
​
​Then, solving the second equation, we get
13x + 5 = −6
13x + 5 − 5 = −6 − 5
13x = −11
​
13x
13
​Therefore the solution to ∣13x + 5∣ = 6​is x =
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​
=−
11
13
x =−
11
13
​
​
​
1
11
​or −
​.
13
13
​
​
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Derivita: SAT PREP Diagnostic test: Linear Equations
SAT PREP Diagnostic test: Linear Equations
13.
Solve the following equation. If there are multiple answers, enter each answer separated by a
comma. Enter None​if there are no solutions.
∣13x + 5∣ = −6
x=
Solution
In general, any equation written in the form ∣ax + b∣ = c​where c &lt; 0​will have no solutions. Since the
given equation is an absolute value expression equal to a negative number, we can conclude there are no
solutions.
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2/3/22, 12:42 PM
Derivita: SAT PREP Diagnostic test: Linear Equations
SAT PREP Diagnostic test: Linear Equations
14.
Given the equation
−12x + 3y = 9
a) Solve for y ​if x = −1​.
y=
b) Solve for y ​in general.
y=
Solution
a) To solve for y ​when x = −1​, plug the given value in for x​, then isolate y ​.
−12x + 3y = 9
−12 (−1) + 3y = 9
12 + 3y = 9
12 − 12 + 3y = 9 − 12
​
​
3y = −3
3
3
y =−
3
3
​
​
y = −1
​ ) To solve for y ​in general, we follow the same process as above, but we work with variables instead of
b
plugging in any values.
−12x + 3y = 9
−12x + 12x + 3y = 9 + 12x
3y = 9 − (−12x)
​
3
9 − (−12x)
y =
3
3
​
​
​
y = 3 + 4x
​
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Derivita: SAT PREP Diagnostic test: Linear Equations
SAT PREP Diagnostic test: Linear Equations
15.
Solve for z ​.
x + y + z + 10 = v
z=
Solution
To isolate z ​, we need to subtract the other terms from both sides of the equation.
x + y + z + 10 = v
x − x + y − y + z + 10 − 10 = v − x − y − 10
​
​
z = v − x − y − 10
​
16.
Solve for x​.
y = 4x + 9xz
x=
Solution
To solve the given equation for x​, we start by rewriting the right-hand side by using the distributive
property and then solving. Doing so, we get
y = 4x + 9xz
y = x (4 + 9z )
y
4 + 9z
y
4 + 9z
​Therefore, the rewritten equation is x =
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​
​
​
=
x (4 + 9z )
4 + 9z
​
​
=x
y
​.
4 + 9z
​
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Derivita: SAT PREP Diagnostic test: Linear Equations
SAT PREP Diagnostic test: Linear Equations
17.
Solve for x​.
4 (x + 3) + 8y = −16
x=
Solution
To isolate x​, we need to distribute, then we can subtract​terms from​both sides of the equation.​​
4 (x + 3) + 8y = −16
4x + 12 + 8y = −16
4x + 12 + 8y − 12 − 8y = −16 − 12 − 8y
4x = −28 − 8y
​Finally, we can divide both sides by 4.​
4x = −28 − 8y
4x
4
​
=
−28 − 8y
4
x =−
​
28
8y
−
4
4
​
​
x = −7 − 2y
​
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