From Arithmetic to Algebra: Additional Practice (Set 1)

Simplify each. Inside the parentheses $x + x = 2x$, so $3(x + x) = 3(2x) = 6x$.

$3(x + x) = 6x$

The others fail: $2 + 2x \neq 4x$, $x \cdot x = x^2 \neq 2x$, and $3(x + x) = 6x \neq 3x$.

Division is not commutative, so look for values where the two sides differ. Try $a = 8$, $b = 2$.

$8 \div 2 = 4, \qquad 2 \div 8 = \frac{1}{4}$

The sides differ, so the claim is false. The other choices use $a = b$, where both sides equal $1$ and hide the difference.

Combine the like terms $7x$ and $-4x$; the constant $2$ stays as it is.

$7x - 4x + 2 = 3x + 2$

This is a conditional equation. Subtract $2n$ from both sides, then add $1$.

$4n - 1 = 2n + 5 \;\Rightarrow\; 2n = 6 \;\Rightarrow\; n = 3$

Check: $4(3) - 1 = 11$ and $2(3) + 5 = 11$.

An expression names a value and carries no equals sign, so you evaluate it by substituting a number for $x$.

$3x + 7 \text{ at } x = 2 \text{ is } 3(2) + 7 = 13$

Only an equation, with an equals sign, is solved, and $3x$ and $7$ are not like terms, so it does not equal $10x$.

Substitute, resolve the parentheses, then multiply.

$4(4 + 1) = 4(5) = 20$

Substitute, then divide before adding.

$\frac{12}{3} + 3 = 4 + 3 = 7$

Distribute the leading minus sign, then combine the like terms.

$-(2x - 5) + x = -2x + 5 + x = -x + 5$

Test values, the way this lesson checks a claim, instead of solving.

$0^2 = 0 = 0 + 0, \qquad 2^2 = 4 = 2 + 2, \qquad 3^2 = 9 \neq 3 + 3 = 6$

The two sides agree at $n = 0$ and $n = 2$ but disagree at $n = 3$. So the claim is not an identity (it fails at $n = 3$) and is not never true (it holds at $n = 0$); it is true for some numbers and false for others. Of the four options, only "true at $n = 0$ and $n = 2$" fits.

The distributive law spreads a factor over a sum.

$a(b + c) = ab + ac$

The others are the associative, commutative, and identity laws.

With only subtraction, work left to right.

$10 - 4 - 3 = 6 - 3 = 3$

A number written next to a variable means multiply, not place the digits side by side.

$5x = 5 \times 2 = 10$

Substituting in parentheses, $5(2) = 10$, keeps the multiplication clear and avoids the trap.