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SAT II Math I : Single-Variable Algebra

Study concepts, example questions & explanations for SAT II Math I

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All SAT II Math I Resources

6 Diagnostic Tests 113 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Solving Equations

Solve for $\small y$ in terms of :

$\small \small 8y+3xy-6=2$

Possible Answers:

$\small \frac{-4}{8+3x}$

$\small \frac{-4}{8-3x}$

$\small \frac{8}{8+3x}$

$\small \frac{8}{8-3x}$

Correct answer:

$\small \frac{8}{8+3x}$

Explanation:

$\small \small 8y+3xy-6=2$

To solve this equation in terms of y, we must isolate it. We start by moving those terms, that do not involve y, to the right hand side. Thus, we add 6 to both sides.

$\small 8y+3xy=8$

Next, since there are no other terms on the left hand side that do not include y, we must factor a y out of them.

$\small y(8+3x)=8$

Now, we can divide to isolate y.

$\small y=\frac{8}{8+3x}$

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Example Question #1 : Solving Equations

Solve for the unknown variable: 6-x=6x-6

Possible Answers:

$\frac{1}{6}$

$\frac{12}{7}$

$\frac{7}{12}$

Correct answer:

$\frac{12}{7}$

Explanation:

To solve 6-x=6x-6 , group like terms on one side of the equal sign.

6-x+(x)=6x-6+(x)

6=7x-6

12=7x

$x=\frac{12}{7}$

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Example Question #2 : Solving Equations

Solve for .

19x+34=15

Possible Answers:

Correct answer:

Explanation:

To isolate the variable in the equation, perform the opposite operation to move all constants on one side of the equation and leaving the variable on the other side of the equation.

19x+34=15

Subtract on both sides. Since is greater than and is negative, our answer is negative. We treat as a normal subtraction.

19x=-19

Divide on both sides. When dividing with a negative number, our answer is negative.

x=-1

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Example Question #1 : Solving Equations

Solve for .

13x+45=-72

Possible Answers:

Correct answer:

Explanation:

To isolate the variable in the equation, perform the opposite operation to move all constants on one side of the equation and leaving the variable on the other side of the equation.

13x+45=-72

Subtract on both sides. When adding with another negative number, just treat as an addition problem and then add a negative sign in front.

13x=-117

Divide on both sides. When dividing with a negative number, our answer is negative.

x=-9

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Example Question #172 : Sat Subject Test In Math I

$\frac{x}{14}+34=62$

Possible Answers:

312

352

392

Correct answer:

392

Explanation:

To isolate the variable in the equation, perform the opposite operation to move all constants on one side of the equation and leaving the variable on the other side of the equation.

$\frac{x}{14}+34=62$

Subtract on both sides.

$\frac{x}{14}=28$

Multiply on both sides.

x=392

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Example Question #3 : Solving Equations

Solve for .

$\frac{x}{-7}+12=-14$

Possible Answers:

182

242

152

Correct answer:

182

Explanation:

To isolate the variable in the equation, perform the opposite operation to move all constants on one side of the equation and leaving the variable on the other side of the equation.

$\frac{x}{-7}+12=-14$

Subtract on both sides. When adding with another negative number, just treat as an addition problem and then add a negative sign in front.

$\frac{x}{-7}=-26$

Multiply on both sides. When multiplying with another negative number, our answer is positive.

x=162

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Example Question #173 : Sat Subject Test In Math I

Solve for .

15x-45=60

Possible Answers:

Correct answer:

Explanation:

To isolate the variable in the equation, perform the opposite operation to move all constants on one side of the equation and leaving the variable on the other side of the equation.

15x-45=60

Add to both sides.

15x=105

Divide on both sides.

x=7

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Example Question #7 : Solving Equations

11x-72=-215

Possible Answers:

Correct answer:

Explanation:

To isolate the variable in the equation, perform the opposite operation to move all constants on one side of the equation and leaving the variable on the other side of the equation.

11x-72=-215

Add on both sides. Since 215 is greater than and is negative, our answer is negative. We treat as a normal subtraction.

11x=-143

Divide on both sides. When dividing with a negative number, our answer is negative.

x=-13

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Example Question #174 : Sat Subject Test In Math I

Solve for .

$\frac{x}{8}-5=-2$

Possible Answers:

-3{}

Correct answer:

Explanation:

To isolate the variable in the equation, perform the opposite operation to move all constants on one side of the equation and leaving the variable on the other side of the equation.

$\frac{x}{8}-5=-2$

Add on both sides. Since is greater than and is positive, our answer is positive. We treat as a normal subtraction.

$\frac{x}{8}=3$

Multiply on both sides.

x=24

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Example Question #175 : Sat Subject Test In Math I

Solve for .

$\frac{x}{-4}-3=-15$

Possible Answers:

Correct answer:

Explanation:

To isolate the variable in the equation, perform the opposite operation to move all constants on one side of the equation and leaving the variable on the other side of the equation.

$\frac{x}{-4}-3=-15$

Add to both sides. Since is greater than and is negative, our answer is negative. We treat as a normal subtraction.

$\frac{x}{-4}=-12$

When multiplying with another negative number, our answer is positive.

x=48

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All SAT II Math I Resources

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SAT II Math I : Single-Variable Algebra

Study concepts, example questions & explanations for SAT II Math I

All SAT II Math I Resources

Example Questions

Example Question #1 : Solving Equations

Example Question #1 : Solving Equations

Example Question #2 : Solving Equations

Example Question #1 : Solving Equations

Example Question #172 : Sat Subject Test In Math I

Example Question #3 : Solving Equations

Example Question #173 : Sat Subject Test In Math I

Example Question #7 : Solving Equations

Example Question #174 : Sat Subject Test In Math I

Example Question #175 : Sat Subject Test In Math I

All SAT II Math I Resources

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