Solving Linear Systems Graphically

In class, we worked through this sequence of questions in random groups at the whiteboards.

If you were away

Work through the questions below in order, on graph paper, before looking at any answers. Each question builds on the one before it. Surprised by what you discover in a given question? That is where the learning happens.

Once you have an answer, expand the answer box to check your work. If a question didn’t go the way you expected, try to explain why before moving on.

When you reach the end, read the summary of key ideas that we built together in class.

TIP

A solution to a linear system is a point that satisfies both equations. A quick way to be sure of your answer is to substitute the point into each equation and confirm that the left side equals the right side.

Question 1

Graph both lines on one grid. Where do the lines cross?

$\begin{array}{rlrl}& \text{A}:& y& =2x-3\\ & \text{B}:& y& =-x+3\end{array}$

Answer

The lines cross at $(2,1)$.

Both equations are already in the form $y=mx+b$, so each line can be drawn from its $y$-intercept and slope. Line A is red; line B is teal.

The lines y = 2x - 3 and y = -x + 3 cross at the marked point (2, 1)

Check: $2(2)-3=1$ and $-(2)+3=1$. The point satisfies both equations.

Question 2

Graph both lines on one grid. Where do the lines cross?

$\begin{array}{rlrl}& \text{A}:& y& =x+1\\ & \text{B}:& 2x-y& =2\end{array}$

Answer

The lines cross at $(3,4)$.

Line B is in standard form. You can graph it from its intercepts ($x$-intercept $1$, $y$-intercept $-2$), or rearrange it first to $y=2x-2$.

The lines y = x + 1 and 2x - y = 2 cross at the marked point (3, 4)

Check: $3+1=4$ and $2(3)-4=2$.

Question 3

Solve the system by graphing. Check your point in both equations.

$\begin{array}{rlrl}& \text{A}:& x+y& =4\\ & \text{B}:& x-y& =-2\end{array}$

Answer

The solution is $(1,3)$.

“Solve” means find the crossing point. Rearranged, line A is $y=-x+4$ and line B is $y=x+2$.

The lines x + y = 4 and x - y = -2 cross at the marked point (1, 3)

Check: $1+3=4$ and $1-3=-2$.

Question 4

Solve the system by graphing.

$\begin{array}{rlrl}& \text{A}:& y& =\frac{1}{2}x+2\\ & \text{B}:& 3x+2y& =12\end{array}$

Answer

The solution is $(2,3)$.

Line A has a fraction slope: from the $y$-intercept, go right 2 and up 1. Line B has $x$-intercept $4$ and $y$-intercept $6$.

The lines y = x/2 + 2 and 3x + 2y = 12 cross at the marked point (2, 3)

Check: $\frac{1}{2}(2)+2=3$ and $3(2)+2(3)=12$.

Question 5

Solve the system by graphing.

$\begin{array}{rlrl}& \text{A}:& y& =2x+2\\ & \text{B}:& 4x-2y& =8\end{array}$

Answer

There is no solution. The lines are parallel.

Rearranged, line B is $y=2x-4$. The two lines have the same slope but different $y$-intercepts, so they never meet.

The parallel lines y = 2x + 2 and y = 2x - 4, which never meet

Question 6

Solve the system by graphing.

$\begin{array}{rlrl}& \text{A}:& y& =2x+2\\ & \text{B}:& 4x-2y& =-4\end{array}$

Answer

There are infinitely many solutions. The lines are coincident: they are the same line, drawn twice.

Rearranged, line B is $y=2x+2$, exactly the same as line A. Every point on the line satisfies both equations.

The lines y = 2x + 2 and 4x - 2y = -4 are the same line, drawn once in red and once dashed in teal on top

Question 7

Line A is $y=2x+2$. Write an equation for a line B that meets line A at:

1. exactly one point

Answer

Any line with a different slope, for example $y=-x+4$.

2. no points

Answer

Any line with slope $2$ and a different $y$-intercept, for example $y=2x-5$.

3. every point

Answer

The same line, perhaps written differently, for example $2y=4x+4$.

Could two lines meet at exactly two points?

Answer

No, never. Through any two points there is only one line, so two lines that share two points must be the same line, and then they share every point.

So a linear system has exactly one solution, no solution, or infinitely many solutions. There are no other possibilities.

Question 8

Without graphing, say whether each system has one solution, no solution, or infinitely many. How can you tell?

1. $y=3x-2$ and $y=3x+1$

Answer

No solution. Same slope, different $y$-intercepts: parallel lines.

2. $y=3x-2$ and $y=2x-2$

Answer

One solution. The slopes are different, so the lines must cross exactly once.

3. $y=\frac{1}{2}x-4$ and $y=\frac{1}{2}x-4$

Answer

Infinitely many solutions. Same slope and same $y$-intercept: the same line.

How can you tell?

Compare the slopes first. If they are different, there is exactly one solution. If they are the same, compare the $y$-intercepts: different means no solution, the same means infinitely many.

Question 9

Without graphing, say whether each system has one solution, no solution, or infinitely many.

1. $x+2y=6$ and $3x-y=4$

Answer

One solution. Rearranged, the lines are $y=-\frac{1}{2}x+3$ and $y=3x-4$. The slopes differ.

2. $2x+4y=8$ and $x+2y=4$

Answer

Infinitely many solutions. Both rearrange to $y=-\frac{1}{2}x+2$. The first equation is just the second one doubled.

3. $3x-y=2$ and $6x-2y=1$

Answer

No solution. Rearranged, the lines are $y=3x-2$ and $y=3x-\frac{1}{2}$. Same slope, different $y$-intercepts.

How can you tell?

When an equation is in standard form, rearrange it to $y=mx+b$ first. Then compare the slopes, and stop as soon as they differ.

Question 10

Consider the system below.

$\begin{array}{rlrl}& \text{A}:& y& =3x+2\\ & \text{B}:& 6x-2y& =k\end{array}$

Which values of $k$ give the system:

1. no solution?

Answer

Any value of $k$ except $-4$.

2. infinitely many solutions?

Answer

$k=-4$.

Why?

Rearranged, line B is $y=3x-\frac{k}{2}$. The slopes already match, so the system can never have exactly one solution. The lines coincide only when the $y$-intercepts match, that is, when $-\frac{k}{2}=2$, so $k=-4$.

Question 11

A group said the system below has no solution, because 6 and 12 are different $y$-intercepts. What went wrong?

$\begin{array}{rlrl}& \text{A}:& 2x+3y& =6\\ & \text{B}:& 4x+6y& =12\end{array}$

Answer

The system actually has infinitely many solutions.

The constant in standard form is not the $y$-intercept. Rearranged, both equations become $y=-\frac{2}{3}x+2$, so the two equations describe the same line.

Question 12

Do these three lines all pass through one point? Find the point, and check without graphing.

$\begin{array}{rl}y& =x+1\\ y& =-x+3\\ y& =3x-1\end{array}$

Answer

Yes. All three lines pass through $(1,2)$.

The first two lines cross where $x+1=-x+3$, so $x=1$ and $y=2$. Then check the third line: $3(1)-1=2$. A solution satisfies every equation, so substituting is all the checking you need.

Summary of key ideas

Here is the summary of the concepts we consolidated together in class, after the whiteboard work:

Solving Linear Systems Graphically

NOTE

If the whiteboard conversation didn’t quite click, or you were away, this is the document to read for a concrete summary of the key ideas: what a solution to a linear system is, how to find one by graphing, and the three cases (one solution, no solution, infinitely many).

Curriculum connection

B1.3

identify the relationship between the slopes of parallel and perpendicular lines, and use this relationship to solve related problems;

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B1.5

represent the equations of lines in different forms (e.g., $y=mx+b$, $Ax+By+C=0$, $Ax+By=D$) and translate between these forms, as appropriate for the context.

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