Arithmetic with Expressions: Additional Practice (Set 1)

Distribute the $-3$, turning $-3 \cdot (-4)$ into $+12$, then combine.

$7x - 3(2x - 4) = 7x - 6x + 12 = x + 12$

Expand each choice. Only the last returns $12x - 8$.

$4(3x - 2) = 12x - 8$

The others give $12x + 8$, $12x - 6$, and $12x - 24$.

Distribute, mind the minus, then combine the $x^2$ and $x$ terms.

$2(x^2 - x) - (x^2 - 3x) = 2x^2 - 2x - x^2 + 3x = x^2 + x$

Work from the inside out.

$3y - (2 - y) = 4y - 2, \quad 6y - (4y - 2) = 2y + 2$

Combine the $ab$ terms and the $a$ terms separately.

$5ab + 3ab = 8ab, \quad -2a + a = -a$

The result is $8ab - a$.

Distribute each product, then combine like terms.

$3(2a - b) + 2(b - a) - a = 6a - 3b + 2b - 2a - a = 3a - b$

Distribute the minus, then combine the $x^2$ and $x$ terms.

$x^2 - 2x^2 + x^2 = 0, \quad 3x - x = 2x$

The $x^2$ terms cancel, leaving $2x$.

Expand each. The first three all give $2x + 6$, but the last does not.

$2(x + 6) = 2x + 12 \neq 2x + 6$

Clear the innermost parentheses first.

$3 - (2x - 1) = 4 - 2x, \quad -2(4 - 2x) = -8 + 4x$

Then $10 - 8 + 4x = 4x + 2$.

The constant term is the term with no variable.

$9 - 4x + x^2 \rightarrow 9$

Combine each kind of term.

$5x^2 - 3x^2 = 2x^2, \quad -2x + 2x = 0, \quad 1 - 4 = -3$

The $x$ terms cancel, leaving $2x^2 - 3$.

Distribute all three groups, then combine.

$3(x - 2y) - 2(2x - 3y) + (x + y) = 3x - 6y - 4x + 6y + x + y = y$

The $x$ terms cancel and the $y$ terms leave $y$.