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Linear Equations in Two Variables

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SATMath800 Algebra 03 — Linear Equations in Two Variables
ALGEBRA · AREA 03

Linear Equations in Two Variables

Turn equations into points, points into lines, and lines back into equations. Learn how slope, intercepts, coordinates, and algebra describe the same linear relationship.

One equation. Many solutions. One line.
Coordinate planeQuadrantsSolutions as pointsGraphing linesSlopeInterceptsTwo pointsPoint + slopeStandard formSAT practice

1. The coordinate plane: a quick visual reset

You saw the coordinate plane in Linear Functions. Now we will use it as a problem-solving tool.

x-axis y-axis origin (0,0) Quadrant I Quadrant II Quadrant III Quadrant IV x > 0, y > 0 x < 0, y > 0 x < 0, y < 0 x > 0, y < 0
The sign of x tells you left or right of the y-axis. The sign of y tells you above or below the x-axis.

Positive x

Move to the right of the y-axis.

Negative x

Move to the left of the y-axis.

Positive y

Move above the x-axis.

Negative y

Move below the x-axis.

SAT habit: Before calculating anything, look at the signs of the coordinates. They often tell you where a point must be located.

2. What does an equation in two variables mean?

Consider:

\(x+y=6\)

Unlike a one-variable equation, this equation has many solutions. Any ordered pair \((x,y)\) that makes the equation true is a solution.

Try a point

\((2,4)\)

Check it:

\(2+4=6\)

So \((2,4)\) is a solution.

Try another point

\((3,4)\)

Check it:

\(3+4=7\ne6\)

So \((3,4)\) is not a solution.

P Q x y
Every point on the line is a solution to the equation.

3. Two points determine a line

Once you know two distinct points on a line, you can draw the line through them.

P₁ = (−2, −1) P₂ = (2, 3)
Two distinct points determine exactly one line.
Important: One point is not enough to determine a unique line. Two distinct points are enough.

4. Find the slope from two points

Slope tells us how much y changes for a given change in x.

\(m=\dfrac{y_2-y_1}{x_2-x_1}\)

For \((-2,-1)\) and \((2,3)\):

1
Change in y:
\(3-(-1)=4\)
2
Change in x:
\(2-(-2)=4\)
3
Rise over run:
\(m=\dfrac{4}{4}=1\)

Positive slope

Move right → move up.

\(m>0\)

Negative slope

Move right → move down.

\(m<0\)

Zero slope

Move right → stay level.

\(m=0\)

5. Two points → equation of the line

Given two points, find the slope first. Then use either point to find the y-intercept.

Example Find the equation through \((2,5)\) and \((6,13)\).
1
Find the slope.
\(m=\dfrac{13-5}{6-2}=2\)
2
Start with slope-intercept form.
\(y=2x+b\)
3
Use one point.
\(5=2*(2)+b\)
\(b=1\)
4
Write and verify.
\(y=2x+1\)
\(2*(6)+1=13\;\checkmark\)
SAT strategy: Two points → slope → intercept → equation → verify.

6. Negative slope: right means down

Suppose a line passes through \((1,8)\) and \((5,2)\).

\(m=\dfrac{2-8}{5-1}=-\dfrac32\)
1
Use slope-intercept form.
\(y=-\dfrac32x+b\)
2
Use \((1,8)\).
\(8=-\dfrac32*(1)+b\)
3
Solve for b.
\(b=\dfrac{19}{2}\)
\(y=-\dfrac32x+\dfrac{19}{2}\)
run down
Negative slope: as x increases, y decreases.

7. Given a slope and one point

Two points are not always necessary. A slope and one point also determine a unique line.

Example: A line has slope 3 and passes through \((2,7)\). Find its equation.
1
Start with the form.
\(y=3x+b\)
2
Substitute the point.
\(7=3*(2)+b\)
3
Solve for b.
\(7=6+b\)
\(b=1\)
4
Equation.
\(y=3x+1\)

8. Is a point on the line?

A point lies on a line exactly when its coordinates make the equation true.

Yes

For \(y=-2x+7\), test \((3,1)\):

\(1=-2*(3)+7=1\;\checkmark\)

No

Test \((4,1)\):

\(1\ne-2*(4)+7=-1\)

9. Intercepts: where a line meets the axes

x-intercept

Set \(y=0\). The point has the form \((a,0)\).

y-intercept

Set \(x=0\). The point has the form \((0,b)\).

Example: Find both intercepts of \(2x+3y=12\).

x-intercept

\(2x+3*(0)=12\)
\(x=6\)
\((6,0)\)

y-intercept

\(2*(0)+3y=12\)
\(y=4\)
\((0,4)\)
x y
The line continues in both directions; the intercepts are where it crosses the axes.

10. Standard form ↔ slope-intercept form

Standard form

\(Ax+By=C\)
\(2x+3y=12\)

Slope-intercept form

\(y=mx+b\)
\(y=-\dfrac23x+4\)
1
Isolate the y-term.
\(3y=-2x+12\)
2
Divide by 3.
\(y=-\dfrac23x+4\)
Same line, different form: Rewriting an equation does not change the relationship it represents.

11. Horizontal and vertical lines

Horizontal line

\(y=a\)

Same y-value everywhere. Slope 0.

Vertical line

\(x=b\)

Same x-value everywhere. Slope is undefined.

y = 4 x = −2
Connection to Functions: \(y=a\) is a function of x. \(x=b\) is not a function of x because one input can correspond to many y-values.

12. The SAT problem-solving map

Given two points

Find slope → find intercept → write equation → verify.

Given slope + one point

Start with \(y=mx+b\) → substitute → solve for b.

Given an equation

Find slope/intercepts → test points → interpret the graph.

Big picture: Points, slope, intercepts, equations, tables, and graphs are different ways of describing the same linear relationship.

13. Original SAT-style practice

Question 1 · Identify a solution

Which ordered pair \((x,y)\) is a solution to \(3x-2y=14\)?

  • A. \((2,-4)\)
  • B. \((4,-1)\)
  • C. \((6,2)\)
  • D. \((0,-7)\)
Answer: B. \(3*(4)-2*(-1)=14\).
Question 2 · Find the slope

A line passes through \((-3,7)\) and \((5,-9)\). What is its slope?

  • A. \(2\)
  • B. \(-2\)
  • C. \(-\dfrac12\)
  • D. \(\dfrac12\)
Answer: B. \(m=\dfrac{-9-7}{5-(-3)}=-2\).
Question 3 · Write the equation

Which equation represents the line passing through \((1,4)\) and \((4,13)\)?

  • A. \(y=3x+1\)
  • B. \(y=3x-1\)
  • C. \(y=\dfrac13x+3\)
  • D. \(y=4x+9\)
Answer: A.
1
Find the slope.
\(m=\dfrac{13-4}{4-1}=\dfrac{9}{3}=3\)
2
Start with slope-intercept form.
\(y=3x+b\)
3
Use one of the points to find b.
\(4=3*(1)+b\)
\(4=3+b\)
\(b=1\)
4
Write the equation.
\(y=3x+1\)
Question 4 · Negative slope

A line has slope \(-\dfrac23\) and passes through \((3,5)\). Which equation represents it?

  • A. \(y=-\dfrac23x+3\)
  • B. \(y=\dfrac23x+3\)
  • C. \(y=-\dfrac23x+7\)
  • D. \(y=\dfrac23x+7\)
Answer: C. \(5=-\dfrac23*(3)+b\), so \(b=7\).
Question 5 · Point on a line

Which point lies on the line \(y=4-\dfrac12x\)?

  • A. \((2,2)\)
  • B. \((4,2)\)
  • C. \((6,2)\)
  • D. \((8,1)\)
Answer: B. \(4-\dfrac12*(4)=2\).
Question 6 · Find an intercept

What is the y-intercept of \(5x+2y=18\)?

  • A. \((0,9)\)
  • B. \((0,18)\)
  • C. \((9,0)\)
  • D. \((18,0)\)
Answer: A. Set \(x=0\), giving \(2y=18\) and \(y=9\).
Question 7 · Rewrite an equation

Which equation is equivalent to \(4x-5y=20\)?

  • A. \(y=\dfrac45x-4\)
  • B. \(y=-\dfrac45x+4\)
  • C. \(y=\dfrac45x+4\)
  • D. \(y=-\dfrac45x-4\)
Answer: A. \(-5y=-4x+20\), so \(y=\dfrac45x-4\).
Question 8 · Horizontal line

A line passes through \((-7,3)\) and has slope 0. Which equation represents it?

  • A. \(x=-7\)
  • B. \(y=-7\)
  • C. \(x=3\)
  • D. \(y=3\)
Answer: D. A slope of 0 gives a horizontal line, so y remains 3.
Question 9 · Context and slope

A tank contains 420 liters at the start. Water is removed at 18 liters per minute. Which equation gives the amount \(W\), in liters, after \(t\) minutes?

  • A. \(W=18t+420\)
  • B. \(W=420t-18\)
  • C. \(W=420-18t\)
  • D. \(W=18-420t\)
Answer: C. The initial value is 420 and the amount decreases by 18 each minute.
Question 10 · Two points and the intercept

A line passes through \((2,-3)\) and \((8,9)\). What is its y-intercept?

  • A. \(-7\)
  • B. \(-3\)
  • C. \(1\)
  • D. \(9\)
Answer: A. The slope is 2. Then \(-3=2*(2)+b\), so \(b=-7\).

14. Coordinate-plane practice

Use the graph as part of the mathematics. Read coordinates, identify intercepts, compare slopes, and connect the picture to an equation.

Question 11 · Read a line through the origin
x y (2, 1)
The line passes through the origin and another visible grid point.

The graph shows a line through the origin and the point (2,1). Which equation represents the line?

  • A. \(y=2*x\)
  • B. \(y=\dfrac12*x\)
  • C. \(y=x+2\)
  • D. \(y=\dfrac12*x+2\)
Answer: B.
1
Read two points from the graph.
\((0,0)\) and \((2,1)\)
2
Find the slope.
\(m=\dfrac{1-0}{2-0}=\dfrac12\)
3
Use the y-intercept.
\(b=0\)
4
Write the equation.
\(y=\dfrac12*x\)
Question 12 · Find the area using the intercepts
2x + 3y = 12 x y
Use the equation to find where the line crosses the axes, then use those intercepts as the base and height of the triangle.

The line is represented by the equation \(2x+3y=12\). What is the area of the triangle formed by the line and the positive x- and y-axes?

  • A. 8
  • B. 10
  • C. 12
  • D. 24
Answer: C.
1
Find the x-intercept.

At the x-intercept, \(y=0\).

\(2x+3*(0)=12\)
\(2x=12\)
\(x=6\)

So the x-intercept is \((6,0)\).

2
Find the y-intercept.

At the y-intercept, \(x=0\).

\(2*(0)+3y=12\)
\(3y=12\)
\(y=4\)

So the y-intercept is \((0,4)\).

3
Use the intercepts as the base and height.

The positive axes are perpendicular, so the triangle has base 6 and height 4.

\(\text{Area}=\dfrac12*6*4\)
4
Calculate.
\(\text{Area}=12\)
Question 13 · Use a graph to find a perpendicular line
(0,1) (2,3) (−1,−2) y = x + 1 x y
The blue line is the given line. The gold point is the point through which the new perpendicular line must pass.

The line shown passes through \((0,1)\) and \((2,3)\). A new line is perpendicular to this line and passes through \((-1,-2)\). Which equation could represent the new line?

  • A. \(y=-x-3\)
  • B. \(y=-x+1\)
  • C. \(y=x-1\)
  • D. \(y=x+3\)
Answer: A.
1
Find the slope of the given line.
\(m=\dfrac{3-1}{2-0}=1\)
2
Find the perpendicular slope.

The negative reciprocal of 1 is \(-1\).

\(m_{new}=-1\)
3
Use the point \((-1,-2)\).
\(-2=-1*(-1)+b\)
\(-2=1+b\)
\(b=-3\)
4
Write the equation.
\(y=-x-3\)
Question 14 · Read the solution from two graphs
2 −1 x y
The dashed guides connect the intersection to the x- and y-axes so you can read both coordinates.

The graph represents a system of two linear equations. What is the solution to the system?

  • A. \((1,-1)\)
  • B. \((2,-1)\)
  • C. \((1,1)\)
  • D. \((-1,2)\)
Answer: B.
1
Find the intersection.

A solution to the system is the point that lies on both lines.

2
Read the coordinates using the dashed guides.
\(x=2\qquad y=-1\)
3
Write the ordered pair.
\((2,-1)\)
Question 15 · Find a missing coordinate
(0,0) (2,1) (6,k) x y
Use the two known points on the line to determine the missing coordinate.

The line passes through \((0,0)\) and \((2,1)\). It also passes through \((6,k)\). What is the value of \(k\)?

  • A. 2
  • B. 3
  • C. 4
  • D. 6
Answer: B.
1
Find the slope.
\(m=\dfrac{1-0}{2-0}=\dfrac12\)
2
Write the equation.

The line passes through the origin, so the y-intercept is 0.

\(y=\dfrac12*x\)
3
Use \(x=6\).
\(k=\dfrac12*6=3\)
Question 16 · Identify the coordinates of the intercepts
x = 5 y = −3 x y
The line passes through the two intercepts. Only the nonzero coordinate of each intercept is labeled.

The graph shows a line whose x-intercept has x-value 5 and whose y-intercept has y-value −3. Which choice gives the coordinates of both intercepts?

  • A. \((5,0)\) and \((0,-3)\)
  • B. \((0,5)\) and \((-3,0)\)
  • C. \((5,-3)\) and \((0,0)\)
  • D. \((-3,0)\) and \((5,0)\)
Answer: A.
1
Recall the x-intercept.

At an x-intercept, the y-coordinate is 0.

\(x=5\quad\Rightarrow\quad(5,0)\)
2
Recall the y-intercept.

At a y-intercept, the x-coordinate is 0.

\(y=-3\quad\Rightarrow\quad(0,-3)\)
3
Match both coordinates.
\((5,0)\quad\text{and}\quad(0,-3)\)

15. Mastery check

Read the graph

Identify quadrants, coordinates, intercepts, and the sign of the slope.

Test a point

Substitute coordinates and decide whether the equation is true.

Build the line

Use two points, or one point and the slope.

Move between forms

Rewrite standard form as slope-intercept form and interpret the result.

The big picture: A linear equation in two variables describes a set of points. Those points form a line. The slope tells how the line changes, while the intercept tells where it crosses an axis.
SATMath800 · Algebra · Linear Equations in Two Variables
Observe → Identify → Decompose → Reason → Solve → Verify

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