Linear Equations: Chapter Test

Strip the constant first, then divide by the coefficient. Subtracting $12$ from both sides leaves $5x = -15$, and dividing both sides by $5$ keeps the sign of the right-hand side.

$5x + 12 = -3 \quad\Longrightarrow\quad 5x = -15 \quad\Longrightarrow\quad x = -3$

Checking in the original equation: $5(-3) + 12 = -15 + 12 = -3$.

The radius is multiplied by $2\pi$, so dividing both sides by that whole factor undoes it in one move. The factor is $2\pi$, not $2$ and then something else.

$C = 2\pi r \quad\Longrightarrow\quad \frac{C}{2\pi} = r$

A circle of radius $3$ has $C = 6\pi$, and $\dfrac{6\pi}{2\pi} = 3$, which is the radius it started from.

Gather the variable on one side in each equation and read what is left. Only the second loses its variable to a false statement about numbers.

$2x + 7 = 2x - 7 \quad\Longrightarrow\quad 7 = -7$

A statement about numbers with no variable left in it cannot be repaired by any choice of $x$, and $7 = -7$ is false, so nothing satisfies that equation. The first collapses to $-6 = -6$, which is true for every $x$; the third gives $x = -1$; the fourth gives $x = 4$.

Multiply every term, both fractions and the constant, by the common multiplier $12$, then expand carefully: the $-3$ reaches both terms of $x - 2$.

$4x - 3(x - 2) = 12 \quad\Longrightarrow\quad 4x - 3x + 6 = 12 \quad\Longrightarrow\quad x = 6$

Checking in the original equation: $\dfrac{6}{3} - \dfrac{4}{4} = 2 - 1 = 1$.

Name one unknown and write everything else in terms of it. With $a$ adult tickets there are $a + 40$ student tickets, and the money each kind brings in is its count times its price.

$12a + 7(a + 40) = 1230$

That equation gives $19a + 280 = 1230$, so $a = 50$ and there were $90$ student tickets: $12(50) + 7(90) = 600 + 630 = 1230$.

A property is named for what was done to BOTH sides, not for what appears in the equation. The $-6$ was removed by adding $6$ to both sides, and the division by $4$ was removed by multiplying both sides by $4$.

$\frac{x}{4} - 6 = 2 \;\xrightarrow{\;+6\;}\; \frac{x}{4} = 8 \;\xrightarrow{\;\times 4\;}\; x = 32$

Each move is the inverse of an operation wrapped around $x$, and the wrapping is undone from the outside in.

The denominator $x - 4$ is zero at $x = 4$, so that value is excluded before any work begins. Multiplying every term by $x - 4$ clears it:

$x = 4 + 3(x - 4) \quad\Longrightarrow\quad x = 3x - 8 \quad\Longrightarrow\quad x = 4$

The only candidate is the excluded value, so it must be thrown away and the equation has no solution at all.

Add $2x$ to both sides, which lands it on the right with its sign reversed, then divide the WHOLE right-hand side by $5$.

$5y - 2x = 20 \quad\Longrightarrow\quad 5y = 2x + 20 \quad\Longrightarrow\quad y = \frac{2x + 20}{5}$

At $x = 5$ the original reads $5y - 10 = 20$, so $y = 6$, and the rearrangement gives $\dfrac{10 + 20}{5} = 6$ as well.

Consecutive odd integers rise in steps of $2$, so with $n$ the smallest they are $n$, $n + 2$ and $n + 4$.

$n + (n + 2) + (n + 4) = 111 \quad\Longrightarrow\quad 3n + 6 = 111 \quad\Longrightarrow\quad n = 35$

The three integers are $35$, $37$ and $39$, which do add to $111$, and the question asked for the largest.

Expand both sides first, remembering that a leading minus flips every term inside its parentheses, so $-(x - 3) = -x + 3$.

$6x - 8 = 5x + 1 - x + 3 = 4x + 4 \quad\Longrightarrow\quad 2x = 12 \quad\Longrightarrow\quad x = 6$

Checking in the original: the left is $2(18 - 4) = 28$, and the right is $30 + 1 - 3 = 28$.

Each side is a single fraction, so each numerator may be multiplied by the denominator across from it. Then expand both brackets in full.

$3(2x - 1) = 5(x + 4) \quad\Longrightarrow\quad 6x - 3 = 5x + 20 \quad\Longrightarrow\quad x = 23$

Checking in the original: the left is $\dfrac{45}{5} = 9$ and the right is $\dfrac{27}{3} = 9$.

Multiply every term by $100$ to clear the decimals, then expand the left and collect the two coefficients on the right; both sides come to the same expression.

$0.6x - 0.6 = 0.6x - 0.6 \quad\Longrightarrow\quad -0.6 = -0.6$

The variable cancels and the statement left behind is true, so every number satisfies the equation. Testing two values confirms it: at $x = 0$ both sides are $-0.6$, and at $x = 10$ both sides are $5.4$.

The $-3$ multiplies both terms inside the parentheses, and $-3 \times (-5) = +15$. Multiplication happens before the subtraction of $3$ from $8$ can be considered, so the $8$ stays where it is.

$8 - 6x + 15 = 4x + 1 \quad\Longrightarrow\quad 23 - 6x = 4x + 1 \quad\Longrightarrow\quad 22 = 10x$

So $x = \dfrac{11}{5}$. Checking: the left is $8 - 3\left(\tfrac{22}{5} - 5\right) = 8 + \tfrac{9}{5} = \tfrac{49}{5}$, and the right is $\tfrac{44}{5} + 1 = \tfrac{49}{5}$.

Let $t$ be the number of hours the express travels. The goods train has then been going for $t + 2$ hours, and drawing level means the two distances are equal.

$90t = 60(t + 2) \quad\Longrightarrow\quad 90t = 60t + 120 \quad\Longrightarrow\quad t = 4$

The question asks for a distance, not a time, so multiply back: the express covers $90 \times 4 = 360$ km, and the goods train covers $60 \times 6 = 360$ km in its six hours, which is the same place.

Clear the denominator, then gather the two terms holding $x$ on one side and factor it out; it stands in two terms, so no single division can reach it.

$yx = x + a \quad\Longrightarrow\quad yx - x = a \quad\Longrightarrow\quad x(y - 1) = a \quad\Longrightarrow\quad x = \frac{a}{y - 1}$

The last division requires $y \neq 1$, and the factoring assumed $a \neq 0$: at $a = 0$ the original reads $y = 1$ for every allowed $x$, so it has either no solution or all of them and no single formula reports it. Testing with $a = 6$ and $x = 2$: the original gives $y = 4$, and $\dfrac{6}{4 - 1} = 2$, the value it started from.

The denominators forbid $x = 2$ and $x = -1$. Each side is a single fraction, so multiply each numerator by the denominator across from it and expand.

$6(x + 1) = 3(x - 2) \quad\Longrightarrow\quad 6x + 6 = 3x - 6 \quad\Longrightarrow\quad 3x = -12$

So $x = -4$, which is neither forbidden value and is therefore kept. Checking: the left is $\dfrac{6}{-6} = -1$ and the right is $\dfrac{3}{-3} = -1$.

Expanding the left gives $6x - 3k$, so the variable terms match on both sides for every $k$ and always cancel. What is left is a statement about the constants alone.

$6x - 3k = 6x - 15 \quad\Longrightarrow\quad -3k = -15 \quad\Longrightarrow\quad k = 5$

At $k = 5$ the equation reads $3(2x - 5) = 6x - 15$, whose two sides are the same expression, so every number satisfies it. At any other $k$ the leftover statement is false and nothing satisfies it.

Multiply every term by $4$, including the term that is not a fraction, then expand and collect.

$(3x - 1) + 4x = 2(x + 7) \quad\Longrightarrow\quad 7x - 1 = 2x + 14 \quad\Longrightarrow\quad 5x = 15$

So $x = 3$. Checking in the original: the left is $\dfrac{8}{4} + 3 = 5$ and the right is $\dfrac{10}{2} = 5$.

Clear the denominator, then gather every term holding $x$ on one side, because $x$ appears in two of them once the right-hand side is expanded.

$ax + b = k(cx + d) \quad\Longrightarrow\quad ax - kcx = kd - b \quad\Longrightarrow\quad x(a - kc) = kd - b$

Dividing gives $x = \dfrac{kd - b}{a - kc}$, which requires $a \neq kc$, and also $ad \neq bc$: when $ad = bc$ the left side is the constant $\dfrac{a}{c}$ wherever it is defined, and the formula returns the very value the denominator forbids. Testing with $a = 3$, $b = 1$, $c = 1$, $d = 2$ and $k = 2$: the formula returns $\dfrac{4 - 1}{3 - 2} = 3$, and at $x = 3$ the original reads $\dfrac{10}{5} = 2$.

The multiplier $x + 2$ is a perfectly good nonzero number at almost every value of $x$, and zero at one of them. Wherever it is zero, both sides of the new equation become zero whatever they were before, so that value satisfies the new equation for free. Take $x + 1 = 4$, which only $x = 3$ satisfies, and multiply both sides by $x + 2$:

$(x + 1)(x + 2) = 4(x + 2)$

At $x = -2$ the left is $(-1)(0) = 0$ and the right is $4(0) = 0$, so $x = -2$ now satisfies it, although $-2 + 1 = 4$ is false. The other three moves are undone by a move of the same kind at every value of $x$, so they change nothing.