Solving Linear Systems by Graphing

A linear system is two (or more) linear equations that are true at the same time. Solving it means finding the point that works for both, which is where their lines cross. You met this in Grade 9; here you’ll review the graphing method, learn to predict how many solutions a system has without drawing anything, and see why you’ll soon want algebraic methods as well.

Key ideas

What a solution is

A solution of a system of two equations in $x$ and $y$ is an ordered pair $(x, y)$ that makes both equations true. On a graph, every point on a line satisfies that line’s equation, so a point on both lines is where they intersect.

For example, $(2, 1)$ is a solution of the system

$$\begin{aligned} y &= 2x - 3 \\ y &= -x + 3 \end{aligned}$$

because $2(2) - 3 = 1$ ✓ and $-2 + 3 = 1$ ✓. A point that works in only one equation is not a solution.

Solving by graphing

1. Graph both lines on the same grid. Use $y = mx + b$ (start at $b$, then use the slope), or plot the two intercepts.
2. Read the coordinates of the point where the lines cross.
3. Check the point in both original equations.
4. State the solution as an ordered pair.

One, none, or infinitely many

Two lines in a plane can meet in only three ways:

Different slopes give one solution; parallel lines give none; the same line gives infinitely many.

The lines are…	Slopes and intercepts	Number of solutions
intersecting	different slopes	exactly one
parallel and distinct	same slope, different $y$-intercepts	none
the same line (coincident)	same slope, same $y$-intercept	infinitely many

So you can predict the number of solutions by writing both equations in the form $y = mx + b$ and comparing $m$ and $b$, without graphing at all. When there are infinitely many solutions, every point on the line is a solution.

The limits of graphing

Graphing shows the whole picture, but it is only as accurate as your drawing. If the lines cross at a point like $\left(\tfrac{5}{3}, \tfrac{8}{3}\right)$, you can only estimate it from a grid. To get exact answers every time, use the algebraic methods: substitution and elimination. Graphing technology such as Desmos can find intersections accurately, and it’s a great way to check your algebra.

Worked examples

Example 1: Two lines in slope–intercept form

Solve the system by graphing.

$$\begin{aligned} y &= 2x - 3 \\ y &= -x + 3 \end{aligned}$$

Solution. Graph each line from its $y$-intercept:

• $y = 2x - 3$: start at $(0, -3)$, then go up $2$ and right $1$, to $(1, -1)$, $(2, 1)$, $(3, 3)$.
• $y = -x + 3$: start at $(0, 3)$, then go down $1$ and right $1$, to $(1, 2)$, $(2, 1)$, $(3, 0)$.

Both lists contain $(2, 1)$, so the lines cross there.

Check in the first equation: $2(2) - 3 = 1$ ✓. Check in the second: $-2 + 3 = 1$ ✓.

The solution is $(2, 1)$.

Example 2: Graphing with intercepts

Solve the system by graphing.

$$\begin{aligned} x + y &= 5 \\ 2x - y &= 4 \end{aligned}$$

Solution. Both equations are in the form $Ax + By = D$, so intercepts are the quickest way to graph them.

• $x + y = 5$: when $x = 0$, $y = 5$; when $y = 0$, $x = 5$. Plot $(0, 5)$ and $(5, 0)$.
• $2x - y = 4$: when $x = 0$, $-y = 4$, so $y = -4$; when $y = 0$, $2x = 4$, so $x = 2$. Plot $(0, -4)$ and $(2, 0)$.

Each line is drawn through its two intercepts. They cross at $(3, 2)$.

The lines appear to cross at $(3, 2)$.

Check: $3 + 2 = 5$ ✓ and $2(3) - 2 = 4$ ✓. The solution is $(3, 2)$.

Example 3: How many solutions?

Without graphing, find the number of solutions of each system.

• (a) $y = 3x + 1$ and $6x - 2y = -2$
• (b) $2x + y = 4$ and $y = -2x - 1$
• (c) $y = \dfrac{1}{2}x + 4$ and $x + y = 1$

Solution. Write each equation in the form $y = mx + b$, then compare.

(a) $6x - 2y = -2$ gives $-2y = -6x - 2$, so $y = 3x + 1$. That’s the same slope and the same intercept as the first equation: the same line, so infinitely many solutions.

(b) $2x + y = 4$ gives $y = -2x + 4$. Both slopes are $-2$, but the intercepts are $4$ and $-1$. The lines are parallel: no solution.

(c) $x + y = 1$ gives $y = -x + 1$. The slopes are $\dfrac{1}{2}$ and $-1$, which are different, so the lines cross once: exactly one solution.

Example 4: When graphing can only estimate

Solve the system $y = x + 1$ and $y = -2x + 6$ by graphing, and check your answer.

Solution. Graph both lines. They cross between grid lines, at roughly $(1.7, 2.7)$.

Check the estimate. In the first equation: $1.7 + 1 = 2.7$ ✓. In the second: $-2(1.7) + 6 = 2.6$, not $2.7$. The estimate is close but not exact.

Since both equations are solved for $y$, set the right sides equal (the comparison method from Grade 9):

$$\begin{aligned} x + 1 &= -2x + 6 \\ 3x &= 5 \\ x &= \frac{5}{3} \end{aligned}$$

Then $y = \dfrac{5}{3} + 1 = \dfrac{8}{3}$. The exact solution is $\left(\dfrac{5}{3}, \dfrac{8}{3}\right)$, about $(1.67, 2.67)$. The graph got you close; the algebra got you the exact answer.

Common mistakes

Checking the point in only one equation. A solution must satisfy both equations. A point that’s on one line only isn’t a solution, so always check both.

Trusting an estimate as exact. When the lines cross between grid lines, your reading is approximate. If the check doesn’t work exactly, solve algebraically, or say clearly that the answer is an estimate.

Saying “no solution” when the lines are the same. If both equations simplify to the same $y = mx + b$, every point on the line works: there are infinitely many solutions, not none.

Comparing coefficients instead of slopes. In $2x + y = 4$ the slope is $-2$, not $2$. Rewrite both equations as $y = mx + b$ before you compare.

Sloppy graphs. A small error in a slope or an intercept moves the intersection a lot. Use a ruler, plot at least two points per line (three is safer), and label each line.

Practice

1. (Warm-up) Decide whether each point is a solution of the system.

• (a) $(3, -1)$ for $x + 2y = 1$ and $2x - y = 7$
• (b) $(1, 2)$ for $y = 3x - 1$ and $x + y = 4$

Solution

(a) $3 + 2(-1) = 1$ ✓ and $2(3) - (-1) = 7$ ✓. Yes, $(3, -1)$ is a solution.

(b) $3(1) - 1 = 2$ ✓, but $1 + 2 = 3 \ne 4$ ✗. No: the point is on the first line only.

2. (Warm-up) Solve by graphing: $y = x + 2$ and $y = -x + 4$.

Solution

Points on $y = x + 2$: $(0, 2)$, $(1, 3)$, $(2, 4)$. Points on $y = -x + 4$: $(0, 4)$, $(1, 3)$, $(2, 2)$. The lines cross at $(1, 3)$.

Check: $1 + 2 = 3$ ✓ and $-1 + 4 = 3$ ✓. The solution is $(1, 3)$.

3. (Core) Solve by graphing, using intercepts: $2x + 3y = 12$ and $x - y = 1$.

Solution

$2x + 3y = 12$ has intercepts $(6, 0)$ and $(0, 4)$. $x - y = 1$ has intercepts $(1, 0)$ and $(0, -1)$. The lines cross at $(3, 2)$.

Check: $2(3) + 3(2) = 12$ ✓ and $3 - 2 = 1$ ✓. The solution is $(3, 2)$.

4. (Core) Without graphing, state the number of solutions of each system. Explain.

• (a) $y = -4x + 2$ and $8x + 2y = 4$
• (b) $3x - y = 5$ and $y = 3x + 2$
• (c) $x = 2$ and $y = -3$

Solution

(a) $8x + 2y = 4$ gives $2y = -8x + 4$, so $y = -4x + 2$. Same line: infinitely many solutions.

(b) $3x - y = 5$ gives $y = 3x - 5$. Both slopes are $3$, with intercepts $-5$ and $2$. Parallel lines: no solution.

(c) A vertical line and a horizontal line always cross once: exactly one solution, $(2, -3)$.

5. (Core) Solve the system $x = -1$ and $y = 2x + 5$.

Solution

Every point on $x = -1$ has $x$-coordinate $-1$. On the second line, when $x = -1$, $y = 2(-1) + 5 = 3$. The lines cross at $(-1, 3)$.

6. (Core) Gym A charges a $40 sign-up fee plus $15 a month. Gym B has no sign-up fee and charges $25 a month.

• (a) Write an equation for the total cost $C$ (in dollars) of each gym after $n$ months.
• (b) Graph both equations and find where they cross. What does that point mean?
• (c) Which gym is cheaper for a full year?

Solution

(a) Gym A: $C = 15n + 40$. Gym B: $C = 25n$.

(b) Gym A’s line starts at $(0, 40)$ and rises $15$ per month; Gym B’s starts at $(0, 0)$ and rises $25$ per month. They cross at $(4, 100)$.

Check: $15(4) + 40 = 100$ ✓ and $25(4) = 100$ ✓. After $4$ months, both gyms have cost $100 in total.

(c) After $4$ months, Gym B’s line is above Gym A’s, because it rises faster. For $12$ months, Gym A costs $15(12) + 40 = 220$ dollars and Gym B costs $25(12) = 300$ dollars, so Gym A is cheaper.

7. (Core) The lines $y = 2x$ and $x + y = 4$ cross between grid lines.

• (a) Estimate the solution from a graph.
• (b) Find the exact solution by substituting $y = 2x$ into the second equation.

Solution

(a) The lines cross at roughly $(1.3, 2.7)$.

(b) Replace $y$ with $2x$: $x + 2x = 4$, so $3x = 4$ and $x = \dfrac{4}{3}$. Then $y = 2 \cdot \dfrac{4}{3} = \dfrac{8}{3}$.

The exact solution is $\left(\dfrac{4}{3}, \dfrac{8}{3}\right)$. Check: $\dfrac{4}{3} + \dfrac{8}{3} = \dfrac{12}{3} = 4$ ✓.

8. (Challenge) For what value of $k$ does the system $y = kx + 3$ and $4x - 2y = 6$ have no solution? Is there any value of $k$ that gives infinitely many solutions?

Solution

$4x - 2y = 6$ gives $-2y = -4x + 6$, so $y = 2x - 3$.

No solution means parallel lines: equal slopes and different intercepts. The intercepts are $3$ and $-3$, which are always different, so $k = 2$ gives no solution.

Infinitely many solutions would need the same intercept too, but $3 \ne -3$. So no value of $k$ gives infinitely many solutions.

9. (Challenge) Find the values of $a$ and $b$ that make the system $ax + 2y = 8$ and $3x + y = b$ have infinitely many solutions.

Solution

Solve both equations for $y$:

$$y = -\frac{a}{2}x + 4 \qquad\qquad y = -3x + b$$

For the same line, the slopes must match and the $y$-intercepts must match:

$$-\frac{a}{2} = -3 \;\Rightarrow\; a = 6 \qquad\qquad b = 4$$

Check: with $a = 6$ the first equation is $6x + 2y = 8$; dividing by $2$ gives $3x + y = 4$, exactly the second equation. ✓