Evaluating and Simplifying Expressions: Additional Practice (Set 1)

Substitute $-5$ for each $x$ in parentheses, then apply the order of operations: exponent first, then the multiplication.

$(-5)^2 + 4(-5) - 1 = 25 - 20 - 1 = 4$

The square $(-5)^2 = 25$ is positive, and $4(-5) = -20$, so the value is $4$.

Distribute each factor, watching that the $-3$ flips the signs of the second group.

$(8x - 12) + (-3x + 15) = 5x + 3$

The $x$-terms give $8x - 3x = 5x$ and the constants give $-12 + 15 = 3$.

Combine the $x^2$-terms with each other and the $x$-terms with each other; the two kinds are unlike.

$(5x^2 + x^2) + (-2x + 6x) = 6x^2 + 4x$

The $x^2$-terms give $6x^2$ and the $x$-terms give $4x$; powers never add when combining like terms.

Substitute $-6$ in parentheses. The expression $-x^2$ squares $x$ first, then takes the opposite.

$-(-6)^2 = -(36) = -36$

The square $(-6)^2 = 36$ is positive, and the leading minus makes the result $-36$.

The minus in front of the parentheses flips both terms inside, then combine like terms.

$6x - 3x + 4 - 2 = 3x + 2$

The $x$-terms give $6x - 3x = 3x$ and the constants give $4 - 2 = 2$.

Substitute both values in parentheses, then apply the order of operations: the exponent, then the multiplications.

$2(3)(4) - (4)^2 = 24 - 16 = 8$

The product $2 \cdot 3 \cdot 4 = 24$ and the square $(4)^2 = 16$, so the value is $8$.

Distribute the $3$, and treat the leading minus as $-1$ over the second group.

$3x + 6y - x + y = 2x + 7y$

The $x$-terms give $3x - x = 2x$ and the $y$-terms give $6y + y = 7y$.

Substitute $-3$ in parentheses, square first, then multiply, then subtract.

$5 - 2(-3)^2 = 5 - 2(9) = 5 - 18 = -13$

The square $(-3)^2 = 9$ is positive, so $2 \times 9 = 18$ is subtracted from $5$.

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

$4x + 2x^2 - 2x + 3x^2 = 5x^2 + 2x$

The $x^2$-terms give $2x^2 + 3x^2 = 5x^2$ and the $x$-terms give $4x - 2x = 2x$.

Substitute $\tfrac{1}{2}$ in parentheses, square first, then multiply and subtract.

$3\left(\tfrac{1}{2}\right)^2 - \tfrac{1}{2} = 3 \cdot \tfrac{1}{4} - \tfrac{1}{2} = \tfrac{3}{4} - \tfrac{2}{4} = \tfrac{1}{4}$

Squaring $\tfrac{1}{2}$ gives $\tfrac{1}{4}$, so $3 \cdot \tfrac{1}{4} = \tfrac{3}{4}$, then subtract $\tfrac{1}{2} = \tfrac{2}{4}$.

Substitute both values in parentheses and square each before adding.

$(-2)^2 + (5)^2 = 4 + 25 = 29$

Both squares are positive: $(-2)^2 = 4$ and $(5)^2 = 25$, so the sum is $29$.

Substitute $3$ for each $x$, finishing the numerator before dividing by the denominator.

$\frac{2(3) + 6}{(3)} = \frac{12}{3} = 4$

The numerator $6 + 6 = 12$ is found first, then divided by $3$ to give $4$.