Linear Equations Practice Questions

Isolate \(x\) by performing the same operation on both sides. First, subtract 7 from both sides:

\(3x = 15\)

Then divide both sides by 3:

\(x = 5\)

Verify: \(3(5) + 7 = 15 + 7 = 22\) ✓

First, distribute the 2 on the left side:

\(2x – 6 + 4 = 3x – 1\)

Combine like terms on the left:

\(2x – 2 = 3x – 1\)

Subtract \(2x\) from both sides:

\(-2 = x – 1\)

Add 1 to both sides:

\(x = -1\)

Verify: \(2(-1 – 3) + 4 = 2(-4) + 4 = -4\) and \(3(-1) – 1 = -4\) ✓

The slope between two points \((x_1, y_1)\) and \((x_2, y_2)\) is:

\(m = \dfrac{y_2 – y_1}{x_2 – x_1}\)

Substitute the given points:

\(m = \dfrac{9 – 3}{4 – 1} = \dfrac{6}{3} = 2\)

Choice D gives only the change in \(y\) (6) without dividing by the change in \(x\). Choice C flips the fraction, computing \(\frac{\Delta x}{\Delta y}\) instead of \(\frac{\Delta y}{\Delta x}\).

This equation is already in slope-intercept form: \(y = mx + b\), where \(m\) is the slope and \(b\) is the \(y\)-intercept.

Comparing \(y = -2x + 5\) with \(y = mx + b\):

• Slope: \(m = -2\)
• \(y\)-intercept: \(b = 5\), which is the point \((0, 5)\)

Choice C swaps the slope and intercept. Choice D gives \((5, 0)\), which is the \(x\)-intercept, not the \(y\)-intercept.

To convert to slope-intercept form \((y = mx + b)\), isolate \(y\). Subtract \(2x\) from both sides:

\(3y = -2x + 12\)

Divide every term by 3:

\(y = -\dfrac{2}{3}x + 4\)

Choice C forgets to divide the constant term (12) by 3. Choice D forgets to divide the \(x\)-coefficient by 3. When dividing both sides by a number, every term must be divided.

Use point-slope form:

\(y – y_1 = m(x – x_1)\)

Substitute \(m = -3\) and the point \((2, 1)\):

\(y – 1 = -3(x – 2)\)

Distribute and solve for \(y\):

\(y – 1 = -3x + 6\)

\(y = -3x + 7\)

Verify: when \(x = 2\), \(y = -3(2) + 7 = 1\) ✓

First, find the slope:

\(m = \dfrac{11 – 5}{3 – 1} = \dfrac{6}{2} = 3\)

Now use point-slope form with \(m = 3\) and either point. Using \((1, 5)\):

\(y – 5 = 3(x – 1)\)

\(y – 5 = 3x – 3\)

\(y = 3x + 2\)

Verify with the other point: \(y = 3(3) + 2 = 11\) ✓

The \(x\)-intercept is the point where the line crosses the \(x\)-axis, which means \(y = 0\). Substitute \(y = 0\) into the equation:

\(4x – 2(0) = 8\)

\(4x = 8 \implies x = 2\)

The \(x\)-intercept is \((2, 0)\).

Choice B gives the \(y\)-intercept (set \(x = 0\) to get \(y = -4\)). Remember: for the \(x\)-intercept, set \(y = 0\); for the \(y\)-intercept, set \(x = 0\).

Parallel lines have the same slope. The given line has slope 4, so the parallel line also has slope 4.

Use point-slope form with \(m = 4\) and the point \((1, 2)\):

\(y – 2 = 4(x – 1)\)

\(y – 2 = 4x – 4\)

\(y = 4x – 2\)

Choice C uses the opposite sign for the slope (−4), and choice D uses the reciprocal of the slope, which would apply to perpendicular lines, not parallel.

Perpendicular lines have slopes that are negative reciprocals of each other. The given line has slope 2, so the perpendicular slope is \(-\frac{1}{2}\).

Use point-slope form with \(m = -\frac{1}{2}\) and the point \((4, 1)\):

\(y – 1 = -\dfrac{1}{2}(x – 4)\)

\(y – 1 = -\dfrac{1}{2}x + 2\)

\(y = -\dfrac{1}{2}x + 3\)

Choice B uses the reciprocal but forgets to negate it. Choice C uses the negative of the slope (−2) rather than the negative reciprocal.

To clear the fractions, multiply every term by the LCD of 6:

\(6 \cdot \dfrac{x}{2} + 6 \cdot \dfrac{x}{3} = 6 \cdot 5\)

\(3x + 2x = 30\)

\(5x = 30\)

\(x = 6\)

Verify: \(\frac{6}{2} + \frac{6}{3} = 3 + 2 = 5\) ✓

Extract the slope from each equation:

• Choice A: \(y = 3x – 1\) → slope = 3
• Choice B: \(2x + y = 5 \implies y = -2x + 5\) → slope = −2
• Choice C: \(y – 2 = 4(x – 1)\) → slope = 4
• Choice D: \(y = x + 4\) → slope = 1

Comparing 3, −2, 4, and 1, the greatest slope is 4 (choice C). Note that you don’t need to fully convert every equation; just identify the coefficient of \(x\).

A horizontal line has a slope of 0 and is parallel to the \(x\)-axis. Every point on a horizontal line shares the same \(y\)-coordinate. Since the line passes through \((3, -7)\), the equation is:

\(y = -7\)

Choice B gives a vertical line through \(x = 3\), not a horizontal line. Remember:

• Horizontal lines: \(y = k\) (slope = 0)
• Vertical lines: \(x = k\) (slope is undefined)

This situation is modeled by a linear equation where the service call fee is the \(y\)-intercept (the starting cost) and the hourly rate is the slope (cost per hour):

\(C(h) = 30h + 50\)

Substitute \(h = 4\) hours:

\(C(4) = 30(4) + 50 = 120 + 50 = 170\)

The total cost is $170. Choice A computes only the labor cost (\(30 \times 4 = 120\)) and forgets the $50 service call fee.

Find the slope of each line:

Line A: \(m_A = \frac{0 – 4}{2 – 0} = \frac{-4}{2} = -2\)

Line B: \(m_B = \frac{2 – 1}{3 – 1} = \frac{1}{2}\)

Since the slopes are negative reciprocals and their product is −1, the lines are perpendicular.