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Linear Equations in Two Variables

What this skill is

A linear equation in two variables, such as y=2x−4y = 2x - 4 or 3x+5y=303x + 5y = 30, describes a line in the xyxy-plane. Every point on the line is a solution. The SAT asks you to move between slope-intercept form and standard form, find intercepts, write equations for parallel or perpendicular lines, and set up equations like 8a+5c=2008a + 5c = 200 from a story.

Key ideas

  • A point (p,q)(p, q) is on a line exactly when substituting x=px = p and y=qy = q makes the equation true.
  • In y=mx+by = mx + b, you can read the slope and yy-intercept directly.
  • In Ax+By=CAx + By = C, the intercepts are quick: set x=0x = 0 to get the yy-intercept and y=0y = 0 to get the xx-intercept.
  • In a model like 8a+5c=2008a + 5c = 200 (adult tickets at $8, child tickets at $5, $200 total), each coefficient is a price per item and the constant is the total. An intercept means "all of one kind": a=25a = 25 when c=0c = 0.
  • Parallel lines have equal slopes. Perpendicular lines have slopes that are negative reciprocals.

Formulas and rules

FormSlopeyy-interceptxx-intercept
y=mx+by = mx + bmmbb−bm-\frac{b}{m}
Ax+By=CAx + By = C−AB-\frac{A}{B}CB\frac{C}{B}CA\frac{C}{A}
  • Point-slope form: y−y1=m(x−x1)y - y_1 = m(x - x_1).
  • Perpendicular slopes multiply to −1-1: slope 25\frac{2}{5} pairs with −52-\frac{5}{2}.

Worked example 1 (easy)

Find the slope and both intercepts of 6x−3y=126x - 3y = 12.

  1. Slope: −AB=−6−3=2-\frac{A}{B} = -\frac{6}{-3} = 2.
  2. yy-intercept: set x=0x = 0, so −3y=12-3y = 12 and y=−4y = -4.
  3. xx-intercept: set y=0y = 0, so 6x=126x = 12 and x=2x = 2.

Check by rewriting: −3y=−6x+12-3y = -6x + 12, so y=2x−4y = 2x - 4. Slope 22, yy-intercept −4-4. Correct.

Worked example 2 (SAT-level)

Line kk is perpendicular to the graph of 3x+2y=83x + 2y = 8 and passes through (6,−1)(6, -1). What is the yy-intercept of line kk?

  1. Slope of the given line: −32-\frac{3}{2}.
  2. Perpendicular slope: the negative reciprocal, 23\frac{2}{3}.
  3. Point-slope form: y+1=23(x−6)y + 1 = \frac{2}{3}(x - 6), so y+1=23x−4y + 1 = \frac{2}{3}x - 4.
  4. Solve for yy: y=23x−5y = \frac{2}{3}x - 5. The yy-intercept is −5-5 (the point (0,−5)(0, -5)).

Check: at x=6x = 6, y=4−5=−1y = 4 - 5 = -1. The line passes through (6,−1)(6, -1), as required.

Common traps

  • Reading the slope of Ax+By=CAx + By = C as AB\frac{A}{B}. The slope is −AB-\frac{A}{B}; in 2x+5y=202x + 5y = 20 it is −25-\frac{2}{5}.
  • Half a negative reciprocal. The perpendicular slope to 25\frac{2}{5} is −52-\frac{5}{2}. Flip the fraction and change the sign.
  • Swapping intercepts. In 4x−3y=244x - 3y = 24, the yy-intercept is 24−3=−8\frac{24}{-3} = -8, not 66 (that is the xx-intercept).
  • Wrong axis in a context. If nn is on the horizontal axis, the horizontal intercept is the value of nn when the other variable is 00.
  • Forgetting to find a constant first. If a line 3x+ky=213x + ky = 21 passes through a given point, plug the point in to get kk before computing anything else.
Practice questions