### All SAT Math Resources

## Example Questions

### Example Question #1 : Algebra

If 6*x* = 42 and *xk* = 2, what is the value of *k*?

**Possible Answers:**

1/7

1/6

5

2/7

7

**Correct answer:**

2/7

Solve the first equation for *x *by dividing both sides of the equation by 6; the result is 7. Solve the second equation for *k *by dividing both sides of the equation by *x, *which we now know is 7. The result is 2/7.

### Example Question #1 : Algebra

If 4*x* + 5 = 13*x* + 4 – *x* – 9, then *x* = ?

**Possible Answers:**

0

8

–5/4

5/4

5/8

**Correct answer:**

5/4

Start by combining like terms.

4*x* + 5 = 13*x* + 4 – *x* – 9

4*x* + 5 = 12*x* – 5

–8*x* = –10

*x* = 5/4

### Example Question #2 : Algebra

If 3 – 3*x* < 20, which of the following could **not** be a value of *x*?

**Possible Answers:**

–5

–4

–2

–3

–6

**Correct answer:**

–6

First we solve for *x*.

Subtracting 3 from both sides gives us –3*x* < 17.

Dividing by –3 gives us *x* > –17/3.

–6 is less than –17/3.

### Example Question #1 : Algebra

Let x be a number. Increasing x by twenty percent yields that same result as decreasing the product of four and x by five. What is x?

**Possible Answers:**

**Correct answer:**25/14

The problem tells us that increasing x by twenty percent gives us the same thing that we would get if we decreased the product of four and x by five. We need to find expressions for these two situations, and then we can set them equal and solve for x.

Let's find an expression for increasing x by twenty percent. We could represent this as x + 20%x = x + 0.2x = 1.2x = 6x/5.

Let's find an expression for decreasing the product of four and x by five. First, we must find the product of four and x, which can be written as 4x. Then we must decrease this by five, so we must subtract five from 4x, which could be written as 4x - 5.

Now we must set the two expressions equal to one another.

6x/5 = 4x - 5

Subtract 6x/5 from both sides. We can rewrite 4x as 20x/5 so that it has a common denominator with 6x/5.

0 = 20x/5 - 6x/5 - 5 = 14x/5 - 5

0 = 14x/5 - 5

Now we can add five to both sides.

5 = 14x/5

Now we can multiply both sides by 5/14, which is the reciprocal of 14/5.

5(5/14) = (14x/5)(5/14) = x

25/14 = x

The answer is 25/14.

### Example Question #2 : Algebra

If 4*xs = v, v = ks , and sv ≠ *0, which of the following is equal to *k *?

**Possible Answers:**

2*xv*

4*xv*

*x*

*xv*

4*x*

**Correct answer:**

4*x*

This question gives two equalities and one inequality. The inequality (* sv ≠ 0*) simply says that neither

*s*nor

*v*is 0. The two equalities tell us that 4

*xs*and

*ks*are both equal to v, which means that 4

*xs*and

*ks*must be equal to each other--that is, 4

*xs*=

*ks*. Dividing both sides by s gives 4

*x*=

*k*, which is our solution.

### Example Question #1 : How To Find The Solution To An Equation

If 2x^{2}(5-x)(3x+2) = 0, then what is the sum of all of the possible values of x?

**Possible Answers:**

**Correct answer:**13/3

Since we are told that 2x^{2}(5-x)(3x+2) = 0, in order to find x, we must let each of the factors of our equation equal zero. The equation is already factored for us, which means that our factors are 2x^{2}, (5-x), and (3x+2). We must let each of these equal zero separately, and these will give us the possible values of x that satisfy the equation.

Let's look at the factor 2x^{2 }and set it equal to zero.

2x^{2 }= 0

x = 0

Then, let's look at the factor 5-x.

5-x = 0

Add x to both sides

5 = x

x = 5

Finally, we set the last factor equal to zero.

(3x+2) = 0

Subtract two from both sides

3x = -2

Divide both sides by three.

x = -2/3

This means that the possible values of x are 0, 5, or -2/3. The question asks us to find the sum of these values.

0 + 5 + -2/3

5 + -2/3

Remember to find a common denominator of 3.

15/3 + -2/3 = 13/3.

The answer is 13/3.

### Example Question #1 : Linear / Rational / Variable Equations

If bx + c = e – ax, then what is x?

**Possible Answers:**

**Correct answer:**(e-c) / (b+a)

To solve for x:

bx + c = e – ax

bx + ax = e – c

x(b+a) = e-c

x = (e-c) / (b+a)

### Example Question #1 : Algebra

√( x^{2} -7) = 3

What is x?

**Possible Answers:**

**Correct answer:**4

To solve, remove the radical by squaring both sides

(√( x^{2} -7))^{ 2} = 3^{2}

x^{2} -7 = 9

x^{2} = 16

x = 4

### Example Question #2 : Algebra

√(3x) = 9

What is x?

**Possible Answers:**

–3

9

–27

27

3

**Correct answer:**

27

To solve, remove the radical by squaring both sides

(√3x)^{ 2} = 9^{2}

3x = 81

x = 81/3 = 27

### Example Question #10 : Algebra

√(8y) + 18 = 4

What is y?

**Possible Answers:**

**Correct answer:**24.5

First, simplify the equation:

√(8y) + 18 = 4

√(8y) = -14

Then square both sides

(√8y)^{ 2} = -14^{2}

8y = 196

y = 196/8 = 24.5

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