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Evaluating and Simplifying Expressions: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 A grouped value

Find the value of $-\frac{(m+n)^2}{4}$ when $m=-5$ and $n=1$ .
1. Hint 1
  The minus sign in front acts on the value of the fraction.
2. Hint 2
  Add the two substituted inputs inside the parentheses, then square that sum before dividing.
Answer

$-4$

Full solution

Substitute both values in parentheses.
The grouped sum is
$(-5)+(1)=-4$

Square the entire sum, then divide by $4$ and apply the outside minus:
$\begin{aligned}(-4)^2&=16\\-\frac{16}{4}&=-4\end{aligned}$

Answer

$-4$

Key idea

Preserve the original grouping and the position of an outside minus during substitution.
Problem 1: Mark complete
Problem 2 A line with letters

Write $6ab-ab+2ab$ with no like terms left to combine.
1. Hint 1
  Compare the complete variable part of each term.
2. Hint 2
  The middle term has coefficient $-1$ , so include that coefficient when counting the copies.
Answer

$7ab$

Full solution

All three terms have the same variable part $ab$ .
Their coefficients are $6$ , $-1$ , and $2$ , so distribution read backward gives
$6ab-ab+2ab=(6-1+2)ab$

Adding the coefficients gives
$6-1+2=7$
Thus the expression is $7ab$ ; the variable part has not changed.

Answer

$7ab$

Key idea

Like terms may contain more than one letter as long as their complete variable parts match.
Problem 2: Mark complete
Problem 3 A penciled expression

Write $-(4p-2q+9)$ without parentheses.
1. Hint 1
  The outside minus applies to the entire grouped expression.
2. Hint 2
  Read the outside sign as multiplication by $-1$ and apply it to each term.
Answer

$-4p+2q-9$ , in any term order.

Full solution

The leading minus means $-1$ times the group.
Its products with $4p$ , $-2q$ , and $9$ are $-4p$ , $2q$ , and $-9$ .
Therefore
$-(4p-2q+9)=-4p+2q-9$

Each term keeps the sign resulting from that multiplication, even if the terms are reordered.

Answer

$-4p+2q-9$ , in any term order.

Key idea

A minus in front of a group changes the sign of every term inside it.
Problem 3: Mark complete
Problem 4 Two envelopes of cards

Two envelopes hold $3r+2$ cards and $2r-5$ cards, where $r$ is a whole number at least $3$ . Five loose cards are then put into the first envelope. Write the total number of cards in the two envelopes with no like terms remaining, then find the total when $r=4$ .
1. Hint 1
  The total includes both envelope counts and the loose cards.
2. Hint 2
  Collect the multiples of $r$ separately from the ordinary numbers.
Answer

$5r+2$ cards; $22$ cards when $r=4$ .

Full solution

Add the three contributions: $3r+2+2r-5+5$ .
The matching variable terms and the constants combine separately:
$\begin{aligned}3r+2r&=5r\\2-5+5&=2\end{aligned}$

The simplified total is $5r+2$ .
Substituting gives
$5(4)+2=22$

The original counts at $r=4$ are $14$ , $3$ , and $5$ cards; their sum is $22$ , which checks the total.

Answer

$5r+2$ cards; $22$ cards when $r=4$ .

Key idea

Combine contributions with matching variable parts and keep the constants in their own group.
Problem 4: Mark complete
Problem 5 A notebook calculation

Write $-(2x^2-7x+4)+3x^2-2x$ with no parentheses or like terms remaining, and find its value when $x=-2$ .
1. Hint 1
  The terms containing $x^2$ are different from those containing $x$ .
2. Hint 2
  Apply the leading minus to every term inside its parentheses.
3. Hint 3
  After combining matching terms, substitute $(-2)$ wherever $x$ appears.
Answer

$x^2+5x-4$ , in any term order; value $-10$ .

Full solution

Removing the parentheses gives $-2x^2+7x-4+3x^2-2x$ .
Collect the squared terms and the first-power terms separately:
$\begin{aligned}-2x^2+3x^2&=x^2\\7x-2x&=5x\end{aligned}$

The constant remains $-4$ , so the result is $x^2+5x-4$ .
At the given input,
$\begin{aligned}(-2)^2+5(-2)-4&=4-10-4\\4-10-4&=-10\end{aligned}$

In the original, the group $2x^2-7x+4$ is $8+14+4=26$ , so with its leading minus it contributes $-26$ .
The term $3x^2$ is $12$ and $-2x$ is $4$ , so the value is
$-26+12+4=-10.$

Answer

$x^2+5x-4$ , in any term order; value $-10$ .

Key idea

After removing parentheses, combine terms by both their letters and their powers.
Problem 5: Mark complete
Problem 6 A bracketed line

Write $4a-[2a-(3-a)]$ with no grouping symbols or like terms remaining.
1. Hint 1
  Work from the innermost grouped expression outward.
2. Hint 2
  Subtracting a group changes the sign of every term in that group.
Answer

$a+3$ , or $3+a$ .

Full solution

First remove the inner parentheses, changing both signs inside, then combine the $a$ terms:
$\begin{aligned}2a-(3-a)&=2a-3+a\\2a-3+a&=3a-3\end{aligned}$

The original expression is now $4a-(3a-3)$ .
Remove that group and combine:
$\begin{aligned}4a-(3a-3)&=4a-3a+3\\4a-3a+3&=a+3\end{aligned}$

As a check at $a=-1$ , the inner group has value $4$ and the brackets have value $-6$ , so the original is $-4-(-6)=2$ , which matches $(-1)+3=2$ .

Answer

$a+3$ , or $3+a$ .

Key idea

With nested groups, remove one group at a time, letting each minus change the sign of every term in the group it stands before.
Problem 6: Mark complete
Problem 7 Two ticket orders

Adult tickets cost $a$ dollars each and child tickets cost $c$ dollars each. Order A is for $4$ adult and $2$ child tickets, every ticket at half price. Order B is for $1$ adult and $3$ child tickets at full price. Write the cost of order A minus the cost of order B as an expression with no parentheses or like terms remaining, then evaluate it when $a=20$ and $c=6$ .
1. Hint 1
  Write each order's cost as its own expression, then subtract order B's whole cost.
2. Hint 2
  Order A's full-price cost is multiplied by $0.5$ , and the minus in front of order B's cost acts as $-1$ .
3. Hint 3
  Both terms in each group must receive the factor outside that group before you collect like terms.
Answer

$a-2c$ dollars, in either term order; $8$ dollars when $a=20$ and $c=6$ .

Full solution

At full price order A's tickets would cost $4a+2c$ dollars, so at half price order A costs $0.5(4a+2c)$ dollars.
Order B costs $a+3c$ dollars, so the difference is
$0.5(4a+2c)-(a+3c)$

Half of $4a$ is $2a$ and half of $2c$ is $c$ , and the minus before the second group makes it $-a-3c$ :
$\begin{aligned}0.5(4a+2c)&=2a+c\\-(a+3c)&=-a-3c\end{aligned}$

Combining matching variable terms in $2a+c-a-3c$ gives
$\begin{aligned}2a-a&=a\\c-3c&=-2c\end{aligned}$

The difference is $a-2c$ dollars.
Substituting the prices gives
$20-2(6)=8$

For a check, order A costs $0.5(80+12)=46$ dollars and order B costs $20+18=38$ dollars, and $46-38=8$ .

Answer

$a-2c$ dollars, in either term order; $8$ dollars when $a=20$ and $c=6$ .

Key idea

Build each amount as its own group, then distribute a decimal factor and a leading minus in the same way before collecting.
Problem 7: Mark complete
Problem 8 Rita's conclusion

Rita rewrites $5r-2r-3r+11$ as $11$ . Her classmate says that because the result is a plain number, Rita must have evaluated the expression rather than simplified it. Are Rita's result and the classmate's objection correct? Explain.
1. Hint 1
  Decide whether $11$ equals the expression for every value of $r$ or only for one chosen value.
2. Hint 2
  Add the coefficients of the three $r$ terms, keeping the sign in front of each.
Answer

Rita's result, $11$ , is correct; the classmate's objection is wrong: Rita simplified.

Full solution

The variable terms can be written as one product:
$5r-2r-3r=(5-2-3)r$

The coefficient is zero, so those terms contribute zero for every $r$ .
Adding the constant gives
$0r+11=11$

The rewrite holds for every value of $r$ , and no value was chosen to reach it, so Rita simplified rather than evaluated.
The classmate's objection fails, because a simplified form can be a plain number when the variable terms cancel.

Answer

Rita's result, $11$ , is correct; the classmate's objection is wrong: Rita simplified.

Key idea

Simplifying can produce a number when the variable terms cancel.
Problem 8: Mark complete
Problem 9 Mark's proposed rewrite

Mark writes $8z^2-3z+z^2$ as $6z^2$ . Is this valid for every value of $z$ ? Give the correct form with no like terms remaining, and test Mark's form and your corrected form at $z=2$ .
1. Hint 1
  Look for terms whose entire variable part matches.
2. Hint 2
  Keep the $z$ term separate from the $z^2$ terms, then compare numerical values.
Answer

No; $9z^2-3z$ ; at $z=2$ the corrected form gives $30$ and Mark's form gives $24$ .

Full solution

The squared terms are alike, but the term $-3z$ has a different power.
Combining the squared terms gives
$8z^2+z^2=9z^2$

The correct simplified form is $9z^2-3z$ .
At $z=2$ , it gives
$9(2)^2-3(2)=30$

Mark's expression gives
$6(2)^2=24$
The original also gives $8(2)^2-3(2)+(2)^2=30$ , so Mark's $24$ differs from the true value and his rewrite is not valid for every input.

Answer

No; $9z^2-3z$ ; at $z=2$ the corrected form gives $30$ and Mark's form gives $24$ .

Key idea

Terms with different powers stay separate when like terms are combined.
Problem 9: Mark complete
Problem 10 Two letters, two reasons

Simplify $2r+5s-7r+3s$ . For each letter, write an equation that shows why its coefficients may be added, and name the property it uses. Do not choose values for the letters.
1. Hint 1
  Only terms with the same variable part can combine, so treat the $r$ terms and the $s$ terms as two separate groups.
2. Hint 2
  When you regroup, the $7r$ term takes the minus sign in front of it, so its coefficient is $-7$ .
3. Hint 3
  In each group the letter is a common factor of both terms; ask which law lets a common factor be pulled out of a sum.
Answer

$-5r+8s$ , or $8s-5r$ ; $2r-7r=(2-7)r$ and $5s+3s=(5+3)s$ , each by the distributive property read backward.

Full solution

Group the $r$ terms and the $s$ terms, keeping the minus sign with $7r$ , so its coefficient is $-7$ .
The expression is now $(2r-7r)+(5s+3s)$ .

In each group the whole variable part is a common factor of both terms, so the distributive property read backward pulls it out of the sum:
$\begin{aligned}2r-7r&=(2-7)r\\5s+3s&=(5+3)s\end{aligned}$

The coefficient sums are $2-7=-5$ and $5+3=8$ , so the expression simplifies to $-5r+8s$ .
Each step used the distributive property, which is true for all numbers, so the rewrite holds for every pair of inputs.

Answer

$-5r+8s$ , or $8s-5r$ ; $2r-7r=(2-7)r$ and $5s+3s=(5+3)s$ , each by the distributive property read backward.

Key idea

Combining like terms is distribution read backward within each matching group.
Problem 10: Mark complete

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