### All SAT Math Resources

## Example Questions

### Example Question #1 : How To Find The Solution For A System Of Equations

If 7*x* + *y* = 25 and 6*x* + *y* = 23, what is the value of *x*?

**Possible Answers:**

20

6

2

11

7

**Correct answer:**

2

You can subtract the second equation from the first equation to eliminate *y*:

7*x* + *y* = 25 – 6*x* + *y* = 23: 7*x* – 6*x* = *x*; *y* – *y* = 0; 25 – 23 = 2

*x* = 2

You could also solve one equation for *y* and substitute that value in for *y* in the other equation:

6*x* + *y* = 23 → y = 23 – 6*x*.

7*x* + *y* = 25 → 7*x* + (23 – 6*x*) = 25 → *x* + 23 = 25 → *x* = 2

### Example Question #1 : How To Find The Solution For A System Of Equations

7x + 3y = 20 and –4x – 6y = 11. Find the value of 3x – 3y

**Possible Answers:**

27

6

16

31

**Correct answer:**

31

We can add these equations to one another.

(7x + 3y = 20) + (–4x – 6y = 11) = (3x – 3y = 31)

### Example Question #3 : How To Find The Solution For A System Of Equations

Consider the three lines given by the following equations:

x + 2y = 1

y = 2x + 3

4x - 3y = 2

What is the value of the x-coordinate of the point of intersection that is common to ALL three lines?

**Possible Answers:**

**Correct answer:**There is no point of intersection

If the point of intersection lies on all three lines, then we should be able to select any two lines, find their point of intersection, and come up with the same point of intersection each time. In other words, the point of intersection of the first two lines must be the point of intersection of the second and third lines.

Let's consider the first and second lines. We can solve the system of equations by substituting the value of y from the second equation into the first.

y = 2x + 3

x + 2(2x + 3) = 1

x + 4x + 6 = 1

5x = -5

x = -1

y = 2(-1) + 3 = 1

The point of intersection of the first two lines is (-1,1).

Now we can find the point of intersection of the second and third lines. Again, we can substitute the value of y from the second equation into the third.

y = 2x + 3

4x - 3(2x + 3) = 2

4x -6x - 9 = 2

-2x = 11

x = -11/2

y = 2(-11/2)+3 = -8

Thus, the second and third lines intersect at (-11/2,-8).

Because the point of intersection between the first and second line does not coincide with the point of intersection between the second and third, there is no point that is common to ALL three lines. Thus, there is no point of intersection.

### Example Question #1 : Systems Of Equations

Each sheep has 4 legs and each chicken has 2 legs. If a farm boy counts 50 heads and 140 feet, how many sheep are there?

**Possible Answers:**

25 sheep

15 sheep

30 sheep

20 sheep

10 sheep

**Correct answer:**

20 sheep

Set x as the number of sheep and y as the number of chicken. This gives us x+y=50 and 4x+2y=140. We want to solve for x. Solving the first equation we get y=50-x. Substitute that into the second you have 4x+2(50-x)=140. Multiplying it out gives 4x+100-2x=140. So 2x+100=140. 2x=40, x=20. Giving 20 sheep.

### Example Question #1 : How To Find The Solution For A System Of Equations

If 8x – 9 is 10 less than 5, what is the value of 4x?

**Possible Answers:**

1

1/4

4

2

1/2

**Correct answer:**

2

The first thing to do is to write an algebrai equation for the problem:

8x – 9 = 5 – 10

8x – 9 = –5

8x = 4

x = 1/2

Thus, 4 * x = 2

### Example Question #2 : How To Find The Solution For A System Of Equations

4x - 5y = 12

6y - 3x = -6

Quantity A: x + y

Quantity B: 6

**Possible Answers:**

Quantity A is greater

The two quantities are equal

The relationship cannot be determined from the information given

Quantity B is greater

**Correct answer:**

The two quantities are equal

Add the two equations:

4x - 5y = 12 plus

6y - 3x = -6:

4x - 5y + (6y - 3x) = 12 + (-6)

4x - 3x + 6y - 5y = 12 - 6

x + y = 6

### Example Question #6 : How To Find The Solution For A System Of Equations

- A charity organization is signing up volunteers to prepare for a fundraiser. Each volunteer can either help setup tables or auction galleries. A volunteer can setup 6 tables per hour or 2 auction galleries per hour. There are 180 tables to be setup as well as 12 auction galleries. If the volunteers will have 3 hours to prepare, how many volunteers must be signed up?

**Possible Answers:**

**Correct answer:**12

Find out how much a volunteer can produce in 3 hours.

6 tables/hour * 3 hours = 18 tables/hour

180 table need to be setup. If one volunteer can setup 18 in 3 hours, then 10 volunteers will take care of the 180 tables.

2 auction galleries/hour * 3 hours = 6 galleries/hour

2 volunteers will be able to complete 12 auction galleries

10 + 2 = 12 volunteers

### Example Question #2 : How To Find The Solution For A System Of Equations

If x + 12 = 28, what is the value of (3x + 2) * (–x + 10)?

**Possible Answers:**

–1300

–180

–300

450

1300

**Correct answer:**

–300

Solve for x, then plug into the formula to find the value. x = 28 – 12 = 16

(3 * 16 + 2) * (–16 +10) = –300

### Example Question #1 : How To Find The Solution For A System Of Equations

Joey has $1.50. If he only has quarters and nickels and he has 14 coins total, how many nickels does he have?

**Possible Answers:**

3

10

6

5

8

**Correct answer:**

10

Setting x and the number of quarters he has and y as the numbver of nickels. x + y = 14 (total coins), 0.25x + 0.05y = 1.50 (total amount). Substituting x = 14 – y from the first equation into the second, we get y = 10. Therefore Joey has 10 nickels.

### Example Question #3 : How To Find The Solution For A System Of Equations

A soccer player kicks a ball at 8m/s. A player runs to receive it as soon as the ball as kicked at a speed of 4m/s. If the receiving player starts 12m ahead of the ball, how far does he travel before he gets the ball?

**Possible Answers:**

9m

3 m

12m

6 m

15m

**Correct answer:**

12m

Setting t as the time elapsed we need to find when 8t = 12 + 4t (this is the distance traveled by the ball compared to the distance traveled by the player+difference from origin). Solving for t we get a travel time of 3 seconds. If the player runs for 3 seconds at 4m/s, the player travels 12m before receiving the ball.

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