From Arithmetic to Algebra: Practice

Distribute the $5$ to each term inside the parentheses.

$5(y + 3) = 5y + 15$

Subtraction is not commutative, so look for values where the two sides differ. Try $a = 5$, $b = 3$.

$5 - 3 = 2, \qquad 3 - 5 = -2$

The sides differ, so $a - b = b - a$ is false. The other choices all use $a = b$, where both sides are $0$ and hide the difference.

The distributive law guarantees the third equation for every $x$.

$2(x + 1) = 2x + 2$

The others hold only for particular values: $x + 2 = 6$ at $x = 4$, $3x = x$ only at $x = 0$, and $x - 3 = 0$ at $x = 3$.

The notation $ab$ means $a \times b$. Substitute and multiply, watching the sign.

$6 \cdot (-2) = -12$

Adding three copies of $x$ is three times $x$.

$x + x + x = 3x$

It is not $x^3$, which would mean $x \cdot x \cdot x$.

Multiply before adding, as the order of operations requires.

$2 + 3(4) = 2 + 12 = 14$

It is not $(2 + 3)(4) = 20$.

The additive inverse adds to the number to give $0$.

$-9 + 9 = 0$

So the additive inverse of $-9$ is $9$.

Factor out the common factor $5$ from both terms.

$10x + 5 = 5(2x + 1)$

Check by distributing: $5 \cdot 2x = 10x$ and $5 \cdot 1 = 5$.

The associative law says the grouping does not change the sum.

$(2 + 5) + 9 = 16, \qquad 2 + (5 + 9) = 16$

Both give $16$, exactly as the law promises for every choice of numbers.

"Decreased by seven" subtracts $7$ from the number, and "equals ten" supplies the equals sign.

$n - 7 = 10$

Because it has an equals sign, it is an equation, not the bare expression $n - 7$.

Three of these are laws that hold for every number; one fails in general because subtraction is not commutative.

$5 - 3 = 2 \neq 3 - 5 = -2$

So $a - b = b - a$ is not an identity, while the commutative, identity, and distributive laws are.

Substitute, then multiply before subtracting.

$12 - 2(4) = 12 - 8 = 4$

It is not $(12 - 2)(4) = 40$; the multiplication comes first.