ACT Math : Algebra

Study concepts, example questions & explanations for ACT Math

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Example Questions

Example Question #1 : How To Evaluate Algebraic Expressions

When graphed in the (x,y) coordinate plane, at what point do the lines -2x + 4y = 5 and y = -2 intersect? 

Possible Answers:

(-13/2,-2)

(2,-2)

(13/2,2)

(13/2,-2)

Correct answer:

(-13/2,-2)

Explanation:

Plugging in y=-2 in the second equation, gives x=-13/2. This is the point where the graphs intersect. 

Example Question #1 : How To Evaluate Algebraic Expressions

The length in cm of a plastic container is 5cm less than triple its width.  Which of the following equations is an accurate description of the length, l, as a function of the width, w?

Possible Answers:

l = 5/3w + 3/5

l = 3w – 5

l = 1/3w + 5

l = 1/3w – 5

l = 3w + 5

Correct answer:

l = 3w – 5

Explanation:

This problem requires the development of an equation.  We are told that the length is 5cm less than 3 times its width.  So we should set up an equation that describes this situation. The equation l = 3w – 5 demonstrates how the length is 5 cm less than 3 times the width of the container.

Example Question #4 : How To Evaluate Algebraic Expressions

The expression x(9 + x)(x – 2) = 4 is a polynomial of which degree?

Possible Answers:

4

1

2

0

3

Correct answer:

3

Explanation:

The highest power this polynomial can achieve is 3.

Example Question #5 : How To Evaluate Algebraic Expressions

Given that x = 2 and y = 3, how much less is the value of  3x2 – 2y than the value of  3y– 2x?

Possible Answers:

29

47

6

1

17

Correct answer:

17

Explanation:

First, we solve each expression by plugging in the given values for x and y:

3(22) – 2(3) = 12 – 6 = 6

3(32) – 2(2) = 27 – 4 = 23

Then we find the difference between the first and second expressions’ values:

23 – 6 = 17

Example Question #6 : How To Evaluate Algebraic Expressions

Evaluate 4x+ 6x – 17, when x = 3. 

Possible Answers:

30

17

13

37

36

Correct answer:

37

Explanation:

Plug in 3 for x, giving you 36 + 18 – 17, which equals 37.

Example Question #7 : How To Evaluate Algebraic Expressions

John has a motorcycle. He drives it to the store, which is 30 miles away. It takes him 30 minutes to drive there and 60 minutes to drive back, due to traffic. What was his average speed roundtrip in miles per hour?

Possible Answers:

30 mph

50 mph

45 mph

60 mph

40 mph

Correct answer:

40 mph

Explanation:

The whole trip is 60 miles, and it takes 90 minutes, which is 1.5 hours.

Miles per hour is 60/1.5 = 40 mph

Example Question #8 : How To Evaluate Algebraic Expressions

If (xy/2) – 3= –9, what is the value of w in terms of x and y?

Possible Answers:

(1/3)xy + 6

w = 3 + (xy/6)

3xy – 6

(1/2)xy – 3

3xy + 6

Correct answer:

w = 3 + (xy/6)

Explanation:

–3w = –9 – (xy/2)

w = 3 + (xy/6)

Example Question #9 : How To Evaluate Algebraic Expressions

Evaluate 5x+ 16x + 7 when x = 7

Possible Answers:

365

363

364

361

362

Correct answer:

364

Explanation:

Plug in 7 for x and you get 5(49) + 16(7) + 7 = 364

Example Question #2 : Evaluating Expressions

Let \dpi{100} \small a\&hash;b=(2a-b)^{2} for all integers \dpi{100} \small a and \dpi{100} \small b. Which of the following is the value of \dpi{100} \small -2\&hash;5?

Possible Answers:

\dpi{100} \small -81

\dpi{100} \small 144

\dpi{100} \small -9

\dpi{100} \small 81

\dpi{100} \small 1

Correct answer:

\dpi{100} \small 81

Explanation:

In order to solve the expression, replace \dpi{100} \small a with \dpi{100} \small -2 and \dpi{100} \small b with \dpi{100} \small 5 in the definition given:

\dpi{100} \small a\&hash;b=(2(-2) -5)^{2}

\dpi{100} \small a\&hash;b=(-4 -5)^{2}

\dpi{100} \small a\&hash;b=(-9)^{2}

\dpi{100} \small a\&hash;b=81

Example Question #11 : How To Evaluate Algebraic Expressions

What is the value of x that satisfies the equation 5(x+2)=12x-5 ?

Possible Answers:

0

-\frac{15}{17}

\frac{15}{7}

1

\frac{15}{11}

Correct answer:

\frac{15}{7}

Explanation:

Distributing the 5 on the left side of the equation gives you 5x+10=12x-5.

Subtracting 5x from both sides of the equation gives you 10=7x-5.

Adding 5 to both sides of the equation gives you 15=7x.

Dividing each side of the equation by 7 gives you \frac{15}{7}=x.

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