From Arithmetic to Algebra: Additional Practice (Set 2)

Simplify each pair. $3(2x) = 6x$, $x + 5 + x = 2x + 5$, and $4x - x = 3x$ all match. But $2$ and $3x$ are not like terms, so they cannot combine into $5x$.

$2 + 3x \neq 5x$

They happen to agree at $x = 1$ ($5 = 5$), but $x = 2$ gives $8 \neq 10$, so they are not equivalent.

Distribute, then combine the like terms.

$5(x + 2) - 3x = 5x + 10 - 3x = 2x + 10$

Substitute, resolve the parentheses first, then multiply.

$3(5 - 8) = 3(-3) = -9$

Substitute in parentheses, square first (the negatives cancel), then subtract.

$(-3)^2 - 4 = 9 - 4 = 5$

The fraction bar groups $x + y$, so add before dividing.

$\frac{7 + 3}{2} = \frac{10}{2} = 5$

Regroup and multiply the numbers first.

$(2 \cdot 5) \cdot n = 10 \cdot n = 10n$

A single successful test does not prove an equation holds for every number, so try another value.

$3^2 = 9 \neq 2(3) = 6$

The equation is true only at $x = 0$ and $x = 2$, so it is a conditional equation, not an identity.

"The product of $4$ and a number" is $4n$, and "increased by twice the number" adds $2n$.

$4n + 2n$

This also equals $6n$ after combining the like terms.

The distributive law guarantees $3(x - 1) = 3x - 3$ for every $x$.

$3(x - 1) = 3x - 3$

The others hold for at most a couple of values: $5x = 5$ only at $x = 1$, $x + x = x^2$ only at $x = 0$ and $x = 2$, and $3(x - 1) = 3x - 1$ never.

Simplify each. Subtracting $x$ from both sides of $x + 1 = x + 2$ leaves $1 = 2$, which is impossible.

$x + 1 = x + 2 \;\Rightarrow\; 1 = 2 \text{ (false)}$

By contrast $x + 2 = 5$ gives $x = 3$, $2x = x$ gives $x = 0$, and $3(x + 1) = 3x + 3$ is an identity, true for every value.

Expand the left side and compare.

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

This would have to equal $2x + 3$, which forces $2 = 3$, impossible. So no value of $x$ works; the equation has no solution.

Distribute both products, then combine the like terms.

$2(3x + 1) + 3(x - 2) = 6x + 2 + 3x - 6 = 9x - 4$