Graph linear equations

1. Every point on the y-axis has x = 0.
2. Substitute x = 0: y = 4(0) - 5 = 0 - 5 = -5
3. So the line crosses the y-axis at (0,\ -5). This matches b = -5 in y = mx + b.

The y-intercept is (0,\ -5).

Check: In y = mx + b, b is the y-intercept; here b = -5 (the sign comes with it) ✓

Tip: In y = mx + b form, you can read the y-intercept straight from the equation: it is (0, b).

A point is on the graph if its coordinates make the equation true. Test each:

1. (4,\ 2): 3(4) - 2 = 10 \ne 2 ✗
2. (2,\ 4): 3(2) - 2 = 6 - 2 = 4 ✓
3. (1,\ 5): 3(1) - 2 = 1 \ne 5 ✗
4. (0,\ 2): 3(0) - 2 = -2 \ne 2 ✗

The point (2,\ 4) is on the line.

Check: Start at the y-intercept (0, -2) and use slope 3: (1, 1), (2, 4) ✓

Tip: "Is this point on the line?" is always a plug-in question: put x in, see if the right y comes out.

1. Points on the x-axis have y = 0. Substitute: 0 = \tfrac{1}{2}x + 4
2. Subtract 4 from both sides: -4 = \tfrac{1}{2}x
3. Multiply both sides by 2 (dividing by \tfrac{1}{2} is the same as multiplying by 2): x = -8
4. So the line crosses the x-axis at (-8,\ 0).

The x-intercept is (-8,\ 0).

Check: \tfrac{1}{2}(-8) + 4 = -4 + 4 = 0 ✓

Tip: To undo "times \tfrac{1}{2}", multiply by 2 — do not multiply by \tfrac{1}{2} again. The point (0,\ 4) is the y-intercept, a different point on the same line.

Test each point in 2x + 5y and look for 20.

1. (2,\ 5): 2(2) + 5(5) = 4 + 25 = 29 ✗
2. (5,\ 2): 2(5) + 5(2) = 10 + 10 = 20 ✓
3. (0,\ 20): 2(0) + 5(20) = 100 ✗
4. (4,\ 0): 2(4) + 5(0) = 8 ✗

The point on the line is (5,\ 2).

Check: Solve for y: y = -\tfrac{2}{5}x + 4. At x = 5: y = -2 + 4 = 2 ✓

Tip: Order matters in (x, y) — the first number always goes with x. Also, the real intercepts here are (10,\ 0) and (0,\ 4): set one variable to 0 and divide by the other coefficient.

1. y-intercept: b = 2, so the line passes through (0,\ 2). Lines A, B and D do; line C crosses at (0, -3), so C is out.
2. Slope: m = -3 = \frac{-3}{1}: from (0, 2) go down 3, right 1 to (1,\ -1). The line falls steeply.
3. Compare: line A passes through (0, 2) and (1, -1) ✓. Line B goes up 3 (slope +3). Line D falls only 1 for every 3 across (slope -\tfrac{1}{3}).

The graph of y = -3x + 2 is line A.

Check: Substitute (1, -1): -3(1) + 2 = -1 ✓

Tip: Use the intercept to eliminate first, then the slope's sign (rising or falling), then its size (steep or gentle).

1. The equation y = -4 has no x in it, so x can be anything while y stays -4.
2. Some points: (-3,\ -4), (0,\ -4), (2,\ -4), (5,\ -4).
3. All of these sit at the same height, 4 units below the x-axis. Joining them gives a horizontal line.
4. It crosses the y-axis at (0,\ -4) and never crosses the x-axis. Its slope is 0 (no rise).

Answer: a horizontal line through (0,\ -4).

Check: Every point listed has y = -4 ✓, and the rise between any two of them is 0, so the slope is 0 ✓

Tip: y = k is horizontal (flat, slope 0); x = h is vertical (undefined slope). Memory trick: "y = number" lines run across, parallel to the x-axis.

1. y-intercept: b = -1, so the line must pass through (0,\ -1). Lines A, B and C pass through (0, -1); line D crosses at (0, 1), so D is out.
2. Slope: m = 2 means from (0, -1) go up 2, right 1 to (1, 1), then to (2, 3).
   • Line A passes through (2, 3): slope \frac{3 - (-1)}{2 - 0} = 2 ✓
   • Line B passes through (-2, 3): slope \frac{3 - (-1)}{-2 - 0} = -2 ✗
   • Line C passes through (4, 1): slope \frac{1 - (-1)}{4 - 0} = \frac{1}{2} ✗

The graph is line A.

Check: On line A at x = 2, y = 3, and 2(2) - 1 = 3 ✓

Tip: Use the intercept to eliminate, then the slope to decide. Rising lines have positive slopes.

1. y-intercept: the line crosses the y-axis at (0,\ 2), so b = 2.
2. Slope: from (0, 2) to (4, -1): m = \frac{-1 - 2}{4 - 0} = \frac{-3}{4} = -\frac{3}{4}
3. Write y = mx + b: y = -\frac{3}{4}x + 2

The equation is y = -\tfrac{3}{4}x + 2.

Check: At x = 4: -\frac{3}{4}(4) + 2 = -3 + 2 = -1 ✓ — the line passes through (4, -1).

Tip: Down 3, right 4 means m = -\frac{3}{4}. Keep "rise" on top and use the sign that matches the direction.

1. y-intercept (move-in fee): at m = 0 the cost is \$40, so b = 40.
2. Slope (monthly fee): from (0, 40) to (4, 180): \frac{180 - 40}{4 - 0} = \frac{140}{4} = 35 \text{ dollars per month}
3. Write C = (\text{slope})\,m + (\text{intercept}): C = 35m + 40

The equation is C = 35m + 40.

Check: At m = 4: 35(4) + 40 = 180 ✓

Tip: In real-world lines, the y-intercept is the starting amount (fee, deposit, initial value) and the slope is the rate (per month, per mile, per hour).

1. The horizontal axis is miles driven; the vertical axis is fare in dollars.
2. At the point (0,\ 4), miles = 0 and fare = 4.
3. So the fare is $4 before any miles are driven — a fixed starting charge.
4. The cost per mile is the slope, not the intercept: m = \frac{14 - 4}{4 - 0} = \frac{10}{4} = 2.5 \text{ dollars per mile}

Answer: a $4 starting charge, paid even for a 0-mile ride.

Check: Equation F = 2.5m + 4. At m = 4: 2.5(4) + 4 = 14 ✓ matches the marked point.

Tip: In real-world line graphs, the y-intercept is the starting value (fee, deposit, first reading) and the slope is the rate (per mile, per hour, per month).

1. Starting value (b): at t = 0, h = 10, so b = 10.
2. Slope (m): use (0, 10) and (8, 0): m = \frac{0 - 10}{8 - 0} = \frac{-10}{8} = -1.25 The candle loses 1.25 inches per hour, so the slope is negative.
3. Write h = mt + b: h = -1.25t + 10

The equation is h = -1.25t + 10.

Check: At t = 4: -1.25(4) + 10 = -5 + 10 = 5 ✓ (the marked point (4, 5)). At t = 8: -10 + 10 = 0 ✓

Tip: When a quantity is used up (burning, draining, paying off), the slope is negative. Rise over run means change in height ÷ change in time, not the other way round.

1. y-intercept: the line crosses the y-axis at (0,\ -3), so b = -3.
2. Slope: from (0, -3) to (2, 2): \text{rise} = 2 - (-3) = 5, \qquad \text{run} = 2 - 0 = 2, \qquad m = \frac{5}{2} The line rises from left to right, so the slope is positive.
3. Write y = mx + b: y = \tfrac{5}{2}x - 3

The equation is y = \tfrac{5}{2}x - 3.

Check: At x = 2: \tfrac{5}{2}(2) - 3 = 5 - 3 = 2 ✓ — the marked point (2, 2).

Tip: Rise is the vertical change (count up/down), run is the horizontal change (count right). Put rise on top. Climbing lines have positive slopes; falling lines have negative slopes.

1. Subtract 3x from both sides: -2y = -3x + 8
2. Divide every term by -2: y = \frac{-3x}{-2} + \frac{8}{-2}
3. Simplify the signs: a negative divided by a negative is positive; a positive divided by a negative is negative. y = \tfrac{3}{2}x - 4

Rosa should get y = \tfrac{3}{2}x - 4 (slope \tfrac{3}{2}, y-intercept -4).

Check: Pick x = 4: from her equation y = 6 - 4 = 2. In the original: 3(4) - 2(2) = 12 - 4 = 8 ✓

Tip: A quick shortcut for Ax + By = C: slope = -\frac{A}{B} = -\frac{3}{-2} = \frac{3}{2}, and y-intercept = \frac{C}{B} = \frac{8}{-2} = -4.

Compute y for x = -1, 0 and 3:

1. x = -1: y = -2(-1) + 5 = 2 + 5 = 7
2. x = 0: y = -2(0) + 5 = 5
3. x = 3: y = -2(3) + 5 = -6 + 5 = -1

The correct table is (-1,\ 7),\ (0,\ 5),\ (3,\ -1).

Check: For a line, the y-values change at a constant rate. From x = 0 to x = 3, y goes from 5 to -1: change -6 over 3 steps = -2 per step ✓ (the slope).

Tip: In a multiple-choice table question, test the hardest point first (a negative or a large x). One wrong point eliminates the whole option.

1. The marked points are (0,\ 300) and (8,\ 900).
2. Change in balance: 900 - 300 = 600 dollars.
3. Change in time: 8 - 0 = 8 weeks.
4. Slope: m = \frac{600}{8} = 75 \text{ dollars per week}
5. In this situation, the slope is how fast the balance grows: Dana deposits $75 each week.

Answer: she deposits $75 per week.

Check: Starting from $300, 8 deposits of $75 add 8 \times 75 = 600, giving $900 ✓

Tip: The units of the slope are (vertical units) per (horizontal unit) — here dollars per week. The $300 is the starting balance (the y-intercept), not the weekly deposit.

1. x-intercept: set y = 0: 3x = 6, so x = 2. Point (2,\ 0).
2. y-intercept: set x = 0: 2y = 6, so y = 3. Point (0,\ 3).
3. Look for the line through (2, 0) and (0, 3): line A.
4. Why not the others? Line B uses (3, 0) and (0, 2) (intercepts swapped). Line C uses (-2, 0) (sign error). Line D uses (0, 6) (forgot to divide 6 by 2).

The graph is line A.

Check: Slope of line A = \frac{0 - 3}{2 - 0} = -\frac{3}{2}, and from the equation y = -\tfrac{3}{2}x + 3 ✓

Tip: In the cover-up method, the x-intercept is C \div A (the x-coefficient) and the y-intercept is C \div B. Matching the wrong coefficient swaps the intercepts.

1. Read the intercepts: x-intercept (6,\ 0) and y-intercept (0,\ -4).
2. Test 2x - 3y = 12 with both points:
   • (6, 0): 2(6) - 3(0) = 12 ✓
   • (0, -4): 2(0) - 3(-4) = 12 ✓
3. Another route: slope = \frac{0 - (-4)}{6 - 0} = \frac{4}{6} = \frac{2}{3}, so y = \frac{2}{3}x - 4. Multiply by 3: 3y = 2x - 12, which rearranges to 2x - 3y = 12.

The equation is 2x - 3y = 12.

Check: The graph rises, and 2x - 3y = 12 has slope -\frac{A}{B} = -\frac{2}{-3} = \frac{2}{3} > 0 ✓

Tip: When answers are in standard form, plugging in the two intercepts is the fastest test — the zeros make the arithmetic easy.

1. Slope: m = \frac{8 - (-1)}{2 - (-4)} = \frac{9}{6} = \frac{3}{2}
2. Find b using (2, 8): 8 = \tfrac{3}{2}(2) + b \;\Rightarrow\; 8 = 3 + b \;\Rightarrow\; b = 5
3. Equation: y = \tfrac{3}{2}x + 5

The line is y = \tfrac{3}{2}x + 5.

Check: The other point: \tfrac{3}{2}(-4) + 5 = -6 + 5 = -1 ✓

Tip: Always check the second point in your final equation — it catches slope and intercept errors in one step. Subtracting a negative (8 - (-1)) means adding.

1. Slope from (-4, 5) to (2, -4): m = \frac{-4 - 5}{2 - (-4)} = \frac{-9}{6} = -\frac{3}{2}
2. y-intercept: the line crosses the y-axis at (0,\ -1), so b = -1. (Check: 5 + (-\tfrac{3}{2})(4) = 5 - 6 = -1 ✓)
3. Slope-intercept form: y = -\tfrac{3}{2}x - 1.
4. Clear the fraction: multiply every term by 2: 2y = -3x - 2.
5. Move the x-term left: add 3x: 3x + 2y = -2.

The equation is 3x + 2y = -2.

Check: (-4, 5): 3(-4) + 2(5) = -12 + 10 = -2 ✓. (2, -4): 6 - 8 = -2 ✓

Tip: When the choices are in standard form, the fastest check is to substitute both marked points into each choice — only one equation works for both.

1. Both points have x = 5. The y-values change, but x does not.
2. So every point on the line has x = 5: the line is vertical, and its equation is x = 5.
3. Slope formula: m = \frac{3 - (-2)}{5 - 5} = \frac{5}{0} Division by zero is not possible, so the slope is undefined.

Answer: The equation is x = 5, and the slope is undefined.

Check: Both points satisfy x = 5 ✓. A horizontal line (y = k) would need equal y-values, which these points do not have.

Tip: Zero slope = flat (horizontal, y = k). Undefined slope = straight up and down (vertical, x = h). "Zero" and "undefined" are not the same thing.

1. Start (y-intercept): at 0 miles the tank holds 14 gallons.
2. Slope: from (0, 14) to (300, 2): m = \frac{2 - 14}{300 - 0} = \frac{-12}{300} = -0.04 \text{ gallons per mile}
3. Interpret: the car uses 0.04 gallon each mile. Flip it to miles per gallon: \frac{1}{0.04} = 25 miles per gallon. So it uses 1 gallon every 25 miles.
4. When empty (x-intercept): 0 = -0.04x + 14 \Rightarrow x = 350 miles — not 300.

Answer: The car started with 14 gallons and uses 1 gallon every 25 miles.

Check: 300 miles at 25 miles per gallon uses 12 gallons; 14 - 12 = 2 gallons left ✓ (the marked point).

Tip: Where the graph stops is not always an intercept. The line ends at (300, 2), but it would reach 0 gallons at 350 miles if extended.