Expressions / Equations

Expressions are groups of mathematical symbols that can be written and manipulated according to definite rules for simplifying or for identifying relationships. For example, (a + 2b)(a – 2b) is an expression. Equations consist of expressions on the two sides of an equals sign, which can be manipulated in order to find the value of one or more variables or “unknowns.” Word problems are usually solved through the formulation and solution of one or more equations.

On this page we provide 12 examples, as follows:

  • 3 Expression problems – General
  • 2 Expression problems with radicals
  • 3 Equation problems – General
  • 2 Simultaneous Linear Equation problems
  • 2 Quadratic Equation problems

To practice more of these types of problems, click here.


Expressions Example 1)

3m + n = x and 2m + 4n = 2x, so what is m/n?

A. 1/3
B. 2
C. 1/2
D. 1/4
E. 1/8

Explanation: Notice that both of these expressions in m and n are equal to some multiple of x. So if we make them equal to the same multiple of x, then we can set them equal to each other and eliminate x. Notice also that we don’t need to find the value of m or n, but just the value of m/n.

3m + n = x and 2m + 4n = 2x

So, 6m + 2n = 2x


6m + 2n = 2m + 4n
6m – 2m = 4n – 2n
4m = 2n
m/n = 2/4 = 1/2

So C is the correct answer.


Expressions Example 2)

If x is an integer, for what value of x is |3x – 28| minimized?

A. -10
B. -9
C. 0
D. 9
E. 10

Explanation: Since the absolute value of any expression is always positive or 0, |3x – 28| is always positive or 0. Since x is an integer 3x is never equal to 28, so the minimum value will come when x = 9 or x = 10. If x = 9, then 3x – 28 = -1, and the absolute value of -1 is 1. If x = 10, then 3x – 28 = 2, and the absolute value of 2 is 2. So |3x – 28| has its minimal value when x = 9.

Therefore, D is the correct answer.


Expressions Example 3)

If uv = mn and vw = nk, which one of the following equalities is valid for all non-zero numbers?

A. uw = nk/v
B. m/u = w/n
C. m/u = k/w
D. w/n = k/m
E. mu = vn


This is not a straightforward problem, so in order to solve it, let’s look at each of the possible answers in turn. Bear in mind that none of these variables is ever equal to 0, but they can be equal to any other number, and they’re not necessarily equal to each other:

A. uw = nk/v

Since we know that vw = nk, then it follows that w = nk/v. Substituting w into answer A for nk/v, we get

uw = w

which is false, for example, if u is 2. So answer A is invalid.

B. m/u = w/n

We know uv = mn, so m/u = v/n. But m/u = w/n implies w = v, which is not necessarily true. So answer B is invalid, also.

C. m/u = k/w

uv = mn means v/n = m/u

vw = nk means v/n = k/w

So, m/u = k/w (Answer C) is valid.

D. w/n = k/m

From vw = nk, we get w/n = k/v. So w/n = k/m implies m = v, which is not necessarily true. So answer D is invalid, also.

E. mu = vn

This implies m = vn/u. Substituting into uv = mn, we get uv = vn2/u, which means u2 = n2, which is not necessarily true. So answer E is also invalid.

Therefore, C is the correct answer.

Notice that we eliminated each incorrect answer by manipulating the answer’s equation and one or more of the two equations we were given originally. We then compared results. We did the same to determine that C was correct.

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Expressions With Radicals Example 1)

\(\begin{align}\sqrt {75} + \sqrt {108} =\end{align}\)

A. \(\begin{align}11\sqrt{3}\end{align}\)
B. \(\begin{align}5\sqrt{3}\end{align}\)
C. \(\begin{align}3\sqrt{5}\end{align}\)
D. \(\begin{align}5\sqrt{5}\end{align}\)
E. \(\begin{align}11\sqrt{7}\end{align}\)

Explanation: Here we have the sum of square roots of two seemingly different numbers. But if we take the prime factorizations of the two and move the factors around a little

\(\begin{align}75 = 3 \cdot 5^2\end{align}\)
\(\begin{align}108 = 3^3 \cdot 2^2 = 3 \cdot 3^2 \cdot 2^2\end{align}\)

we can see that each of these numbers contains a square times the number 3. So if we take their square roots, we get

\(\begin{align}5\sqrt{3} + 6\sqrt{3} = 11\sqrt{3}\end{align}\)

So A is the correct answer.

Note that the important thing to do here is to take the prime factorization of the two numbers. Then we could see how to proceed further.


Expressions With Radicals Example 2)

\(\begin{align}\frac{1}{\sqrt{2}}\, – \sqrt{8} + \frac{\sqrt{72}}{4} =\end{align}\)

A. \(\begin{align}2\end{align}\)
B. \(\begin{align}18\sqrt{2}\end{align}\)
C. \(\begin{align}3\sqrt{5}\end{align}\)
D. \(\begin{align}9\end{align}\)
E. \(\begin{align}0\end{align}\)

Explanation: This problem requires some prime factorization, and other manipulation of radicals for its solution:

\(\begin{align}8 = 2^2 \cdot 2\end{align}\)
\(\begin{align}72 = 3^2 \cdot 2^2 \cdot 2\end{align}\)

When we take the square roots of these two numbers, we get

\(\begin{align}\sqrt{8} = 2\sqrt{2}\end{align}\)
\(\begin{align}\sqrt{72} = 6\sqrt{2}\end{align}\)

We also need to “rationalize” the denominator for the first term. This is done by multiplying this fraction by 1 in the form of \(\begin{align}\frac{\sqrt{2}}{\sqrt{2}} :\end{align}\)

\(\begin{align}(\frac{1}{\sqrt{2}}) \cdot (\frac{\sqrt{2}}{\sqrt{2}}) = \frac{\sqrt{2}}{2}\end{align}\)

Now we are ready to solve the problem:

\(\begin{align}\frac{1}{\sqrt{2}}\, – \sqrt{8} + \frac{\sqrt{72}}{4} =\end{align}\)

\(\begin{align}\frac{\sqrt{2}}{2}\, – {2}\sqrt{2} + \frac{{6}\sqrt{2}}{4} =\end{align}\)

\(\begin{align}\frac{\sqrt{2}}{2}\, – {2}\sqrt{2} + \frac{{3}\sqrt{2}}{2}\end{align}\)

Now we rearrange terms, and we get

\(\begin{align}\frac{\sqrt{2}}{2}\, – {2}\sqrt{2} + \frac{{3}\sqrt{2}}{2} =\end{align}\)

\(\begin{align}\frac{{3}\sqrt{2}}{2} + \frac{\sqrt{2}}{2}\, – {2}\sqrt{2} = \end{align}\)

\(\begin{align}\frac{{4}\sqrt{2}}{2}\, – {2}\sqrt{2} = \end{align}\)

\(\begin{align}{2}\sqrt{2}\; – {2}\sqrt{2} = 0\end{align}\)

So E is the correct answer.

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Equations Example 1)

Today Ralph and his 3 triplet sisters celebrate their birthday. He is 11, and they are 5. In how many years will the combined age of the triplets be twice Ralph’s age?

A. 5
B. 7
C. 8
D. 9
E. 10

Explanation: In word problems we normally use a variable to represent the quantity we are seeking. So we let x equal the number of years until the combined ages of the triplets will be twice Ralph’s age. That being the case, we can use the information we’re given in the problem to formulate an equation:

2(11 + x) = 3(5 + x)

In x years, Ralph will be 11 + x years old, and twice his age at that time will be 2(11 + x). The triplets will be 5 + x years old, and since we’re combining the triplets’ ages, we multiply 5 + x by 3. We then set these two quantities equal to each other, and that gives us the formula for when their combined age will be twice Ralph’s.

Now we solve the equation for x:

22 + 2x = 15 + 3x
22 – 15 = 3x – 2x
7 = x

In 7 years Ralph will be 18 and the triplets will be 12. Since 3 · 12 =36, this is twice Ralph’s age 7 years from now.

So answer B is correct.


Equations Example 2)

21 coins, each worth 5, 10, or 25 cents, add up to $2.15. If there are 2 more dimes (10 cent coins) than nickels (5 cent coins), how many quarters (25 cent coins) are there?

A. 2
B. 3
C. 5
D. 6
E. 8

Explanation: This problem involves 3 unknowns – the numbers of quarters (q), dimes (d), and nickels (n) – and such problems tend to be messy. We’ll find it easier to solve this by picking various possibilities for the number of quarters, and then seeing how each one works. Since we’ve only got 5 possible answers, we’ll try them until we find the right one. Let’s start with answer A – 2 quarters.

If q = 2, then since there are 21 coins altogether

d + n = 19

But d = n + 2, so

n +2 + n = 19
2n = 17
n = 8 1/2

which is absurd, since we can’t have half a nickel. So answer A is out.

In fact, the same thing will happen whenever q is an even number, so we’ve already eliminated all the answers except B and C. Let’s try B.

If q = 3, then

d + n = 18
n + 2 + n = 18
2n = 16
n = 8
d = 10

Then, since q = 3 represents the quarters, d = 10 the dimes, and n = 8 the nickels, by adding the values of these coins together, we get

3∙25 + 10∙10 + 8∙5 = 75 + 100 + 40 = 215

which is the amount they’re supposed to add up to, so the correct answer is B.

What would happen if we chose C, in which q = 5?


n + d = 16
2n + 2 = 16
n = 7
d = 9

In this case then,

5∙25 + 9∙10 + 7∙5 = 125 + 90 + 35 = 250

so the coins add up to too large an amount. Therefore, answer C is incorrect.


Equations Example 3)

MyRetail estimated revenue of $768 on sale of its inventory of a certain product. But when 8 more were found in stock, the store decided to lower its price by $8 each, and in selling all the product, they still received $768, as planned. What was the price they originally planned to charge?

A. $32
B. $24
C. $36
D. $28
E. $30

Explanation: This problem can be solved with equations, but in this case it is easier and faster to simply plug in values from the list of answers, and see what works. Notice that the values of answers C, D, and E do not divide 768 evenly, and so these answers must be incorrect. So let’s try answer A:

768 / 32 = 24

which means that the original quantity was 24. But they found 8 more, so now the quantity is 32, and since they reduced the price by $8, the new price is $24. Thus, the original price was $32, and answer A is correct.

To practice more of these types of problems, click here.


Simultaneous Linear Equations Example 1)

Seamus loves professional sports. In the past 3 years he’s attended 180 baseball, basketball, and soccer games. He’s attended 4 times as many basketball games as soccer games, and 90 more baseball games than basketball games. How many baseball games has he gone to?

A. 50
B. 30
C. 62
D. 45
E. 130

Explanation: Let b be the number of baseball games, k the number of basketball games, and s the number of soccer games Seamus has gone to in the past 3 years. Then we have the following relationships:

b + k + s = 180
k = 4s
b = k + 90

b = 4s + 90
b + k + s = (4s + 90) + 4s + s = 180
9s = 180 – 90 = 90
s = 10
k = 40
b = 130

So answer E is correct.


Simultaneous Linear Equations Example 2)

200 children at a school carnival were each given 1 blue, 1 red, and 1 green chip for use, respectively, at the fishing booth, the merry-go-round, and the throwing gallery. Afterward it was determined that 80% of the blue, 75% of the red, and 40% of the green chips had been used. If none of the children had three chips left over, and 20% of the children had one chip left over, how many had two chips left over?

A. 80
B. 60
C. 85
D. 75
E. 55

Explanation: One difficulty with this problem is caused by the description that’s given. When it says a child had x (0, 1, 2, or 3) chips left over, it really means that the child used 3 – x of the chips. So, let n be the number of children who used 1 chip, or in other words, had two chips left over. Let w be the number that used 2 chips, and let r be the number that used all three chips. Then we need to find the value of n. We have the following relationships:

n + w + r = 200 (since every child used at least one chip)
w = .2∙200 = 40 (since 20% of the children used two chips)

n + 40 + r = 200
n + r = 160

The number of chips used (or spent) was

80% of the blue
75% of the red
40% of the green

and since there were 200 of each color chip, then the total number of chips spent was

(.8 + .75 + .4) x 200 = 390

Now we need a formula showing how the chips were spent. Since 390 chips were spent in all, then that total number can be calculated by adding the number of chips spent by n children, w children, and r children:

n + 2w + 3r = 390

This formula follows from the fact that the w children who spent 2 chips contribute 2w to the total of 390 chips spent, and the r children contribute 3r to the number spent. Since w = 40, then

n + 80 + 3r = 390
n + 3r = 310

We know from above that

n + r = 160

so subtracting this equation from the one above it, we get

2r = 150
r = 75

So finally of the 200 children:

40 children used two chips
75 used three chips
85 used one chip

That is, n = 85.

So C is the correct answer, as the children using one chip (85) still had two left over.

To practice more of these types of problems, click here.


Quadratic Equations Example 1)

\(\begin{align}(p + 2) (p\, – 2) = 2p\, – 1\end{align}\)and\(\begin{align}(q + 2) (q\, – 2) = 2q\, – 1\end{align}\), also\(\begin{align}p > q.\end{align}\)What is the value of\(\begin{align}q\end{align}\)?

A. \(\begin{align}2\end{align}\)
B. \(\begin{align}-1\end{align}\)
C. \(\begin{align}4\end{align}\)
D. \(\begin{align}-2\end{align}\)
E. \(\begin{align}1\end{align}\)

Explanation: These equations are identical, except for the variable in each of them. Additionally, the left side of both is the factoring of the difference of two squares. Therefore, each of these is the same quadratic equation, which will have two solutions. We will find the two solutions, and set one of them equal to p and the other equal to q. We start by expanding each equation so we get:

\(\begin{align}p^2 – 4 = 2p\, – 1\end{align}\)
\(\begin{align}q^2 – 4 = 2q\, – 1\end{align}\)

Now, let’s put everything on the left side and make the right side equal 0, so that these are in the normal form for quadratic equations. We will then factor them so that we can see their solution:

\(\begin{align}p^2 – 2p\, – 3 = 0\end{align}\)
\(\begin{align}(p\, – 3) (p + 1) = 0\end{align}\)
\(\begin{align}p = 3,\end{align}\)or \(\begin{align}p = -1\end{align}\)

\(\begin{align}q^2 – 2q\, -3 = 0\end{align}\)
\(\begin{align}(q\, -3) (q + 1) = 0\end{align}\)
\(\begin{align}q = 3,\end{align}\)or \(\begin{align}q = -1\end{align}\)

Since \(\begin{align}p > q,\end{align}\) then \(\begin{align}p = 3\end{align}\) and \(\begin{align}q = -1.\end{align}\)

So B is the correct answer.


Quadratic Equations Example 2)

If \(\begin{align}a^2b^4 -5 = 4ab^2,\end{align}\)then what is one value of\(\begin{align}a\end{align}\)in terms of\(\begin{align}b\end{align}\)?

A. \(\begin{align}\frac{-1}{5b^2}\end{align}\)
B. \(\begin{align}4b^2\end{align}\)
C. \(\begin{align}5b^2\end{align}\)
D. \(\begin{align}\frac{5}{b^2}\end{align}\)
E. \(\begin{align}-b^2\end{align}\)

Explanation: We can think of this problem as a quadratic equation in\(\begin{align}a,\end{align}\)where the various powers of\(\begin{align}b\end{align}\)are coefficients. So:

\(\begin{align}a^2b^4 – 5 = 4ab^2\end{align}\)
\(\begin{align}a^2b^4 – 5 -4ab^2 = 0\end{align}\)
\(\begin{align}b^4a^2 – 4b^2a\, – 5 = 0\end{align}\)

The expression on the left side of the equals sign can be factored into

\(\begin{align}(b^2a\, – 5) (b^2a + 1) = 0\end{align}\)


\(\begin{align}b^2a\, – 5 = 0,\end{align}\)
\(\begin{align}a = \frac{5}{b^2}\end{align}\)


\(\begin{align}b^2a + 1 = 0\end{align}\)
\(\begin{align}a = \frac{-1}{b^2}\end{align}\)

are the two possible solutions for\(\begin{align}a\end{align}\)in terms of\(\begin{align}b.\end{align}\)The only one of these that shows up in the answers is\(\begin{align}\frac{5}{b^2},\end{align}\)which is answer D.

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