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Graphing Linear Equations: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Table spaces

The table for $x+2y=6$ has three rows: $x=-2$ , $y=a$ ; $x=b$ , $y=2$ ; and $x=4$ , $y=c$ . Complete the table.
1. Hint 1
  Each row must make the equation true.
2. Hint 2
  For the middle row, insert its known $y$ -value; for the other rows, isolate $y$ .
Answer

$a=4$ , $b=2$ , $c=1$ .

Full solution

Solving for $y$ gives
$y=\frac{6-x}{2}$
At $x=-2$ this gives $a=4$ , and at $x=4$ it gives $c=1$ .

The middle row gives
$b+2(2)=6$
Thus $b=2$ .

Answer

$a=4$ , $b=2$ , $c=1$ .

Key idea

Every row in a table of solutions must satisfy the equation, whichever coordinate is missing.
Problem 1: Mark complete
Problem 2 On or off the line

Which of the points $(-1,-8)$ , $(2,1)$ , and $(0,5)$ lie on the graph of $3x-y=5$ ? For each point, show the check that decides it.
1. Hint 1
  The graph is the set of solutions, so each point is tested against the equation itself.
2. Hint 2
  Substitute each point's $x$ and $y$ into $3x-y$ and compare the result with $5$ , taking care with the signs.
Answer

$(-1,-8)$ and $(2,1)$ lie on the graph; $(0,5)$ does not.

Full solution

For $(-1,-8)$ ,
$3(-1)-(-8)=-3+8$
This equals $5$ , so $(-1,-8)$ is on the graph.

For $(2,1)$ ,
$3(2)-1=5$
So $(2,1)$ is on the graph.

For $(0,5)$ ,
$3(0)-5=-5$
This is not $5$ , so $(0,5)$ is off the graph.

Answer

$(-1,-8)$ and $(2,1)$ lie on the graph; $(0,5)$ does not.

Key idea

Substituting a point's coordinates decides whether it is on a line, with no drawing needed.
Problem 2: Mark complete
Problem 3 One drawn line

On the blank grid in the figure, draw the graph of $x-y=x+2$ .

A blank coordinate grid.

Text description of this figure

A blank coordinate grid. The horizontal x-axis runs from negative five to five and the vertical y-axis from negative four to four, with gridlines, tick marks and number labels at every whole number, equal unit lengths on both axes, arrowheads at both ends of each axis, and the origin labeled 0. Nothing is plotted: there are no points, lines or coordinates on the grid.
1. Hint 1
  First decide which coordinate the equation restricts.
2. Hint 2
  Subtract $x$ from both sides and solve the remaining equation.
Answer

The horizontal line $y=-2$ .

Full solution

Subtracting $x$ leaves
$-y=2$
Hence
$y=-2$
Every first coordinate is allowed, while the second coordinate stays $-2$ .
Draw the horizontal line through $(0,-2)$ with arrows at both ends.
Substituting any $x$ gives $x+2$ on both sides.

Answer

The horizontal line $y=-2$ .

Key idea

A variable that cancels from an equation is free to vary along its graph.
Problem 3: Mark complete
Problem 4 Printer record

A printer records $A(-2,8)$ and $C(4,-4)$ as solutions of $2x+y=4$ . Choose an integer $x$ -value strictly between $-2$ and $4$ , and find the matching $y$ -value. On the blank grid in the figure, plot A and C and draw their line, then use your new point to check it.

A blank coordinate grid for plotting $A$ and $C$ .

Text description of this figure

A blank coordinate grid. The horizontal x-axis runs from negative three to five and the vertical y-axis from negative five to nine, with gridlines, tick marks and number labels at every whole number, equal unit lengths on both axes, arrowheads at both ends of each axis, and the origin labeled 0. Nothing is plotted: there are no points, lines or coordinates on the grid.
1. Hint 1
  Solve for $y$ before choosing a new $x$ -value.
2. Hint 2
  Compute the third point from the equation and see whether it lies on the line through the first two.
Answer

Any one of $(-1,6)$ , $(0,4)$ , $(1,2)$ , $(2,0)$ , $(3,-2)$ ; each lies on the line through $A$ and $C$ , the graph of $y=4-2x$ .

Full solution

Solving for $y$ gives
$y=4-2x$
Choose, for example, $x=1$ .
Then
$y=4-2(1)$ $y=2$
Thus one possible third point is $(1,2)$ .
The other allowed choices give $(-1,6)$ , $(0,4)$ , $(2,0)$ and $(3,-2)$ , each an equally good check point.

The endpoint checks are $2(-2)+8=4$ and $2(4)-4=4$ .
Plot A and C and extend the straight line through them.
The third point $(1,2)$ also lies on that line and gives $2(1)+2=4$ , checking the drawing.

Answer

Any one of $(-1,6)$ , $(0,4)$ , $(1,2)$ , $(2,0)$ , $(3,-2)$ ; each lies on the line through $A$ and $C$ , the graph of $y=4-2x$ .

Key idea

Solving for $y$ lets you make extra solutions that check a line drawn through two points.
Problem 4: Mark complete
Problem 5 Supply choices

A shop sells packets for 3 dollars each and rolls for 4 dollars each. Let $x$ count packets and $y$ count rolls in a purchase costing exactly 24 dollars. A purchase may include none of one item. Find both axis crossings of the equation for the cost and draw its full line on the blank grid in the figure. Which points on this line represent possible purchases?

A blank coordinate grid for the cost equation.

Text description of this figure

A blank coordinate grid. The horizontal x-axis runs from negative one to ten and the vertical y-axis from negative two to eight, with gridlines, tick marks and number labels at every whole number, equal unit lengths on both axes, arrowheads at both ends of each axis, and the origin labeled 0. Nothing is plotted: there are no points, lines or coordinates on the grid.
1. Hint 1
  The two costs add to the total.
2. Hint 2
  An axis crossing has one coordinate zero; a purchase needs whole-number counts, so test the whole-number values of $x$ .
Answer

$(8,0)$ and $(0,6)$ ; graph $3x+4y=24$ . Possible purchases: $(0,6)$ , $(4,3)$ , $(8,0)$ .

Full solution

The equation is
$3x+4y=24$
With $y=0$ , $x=8$ ; with $x=0$ , $y=6$ .
Plot $(8,0)$ and $(0,6)$ and extend their line.

The packet count is a whole number from $0$ through $8$ , since $3x$ cannot exceed $24$ , and the roll count is $(24-3x)/4$ .
Checking $x=0$ through $8$ , only $0$ , $4$ and $8$ make $24-3x$ a multiple of $4$ .
The corresponding roll counts are $6$ , $3$ , and $0$ .
Each purchase costs 24 dollars.

Answer

$(8,0)$ and $(0,6)$ ; graph $3x+4y=24$ . Possible purchases: $(0,6)$ , $(4,3)$ , $(8,0)$ .

Key idea

The full graph may include solutions that do not fit the whole-number restrictions of a situation.
Problem 5: Mark complete
Problem 6 Straight up or down

The figure shows points $P$ and $Q$ . Read their coordinates and decide, for each, whether it lies on the graph of $x+2y=7$ . Then move any point that is not on the graph straight up or down until it lands on the graph. Give where it lands and the equation of the vertical line it moves along.

Points $P$ and $Q$ on the coordinate plane.

Text description of this figure

A coordinate grid. The horizontal x-axis and the vertical y-axis each run from negative two to five, with gridlines, tick marks and number labels at every whole number, equal unit lengths on both axes, arrowheads at both ends of each axis, and the origin labeled 0. Two points are plotted and labeled with their letters only: point P, one unit left of the y-axis and four units above the x-axis, at negative one, four; and point Q, one unit right of the y-axis and one unit below the x-axis, at one, negative one. No coordinates, lines or other points are shown.
1. Hint 1
  A point is on the graph exactly when its coordinates make the equation true.
2. Hint 2
  A vertical move keeps the first coordinate fixed; substitute it into the equation.
Answer

$P(-1,4)$ lies on the graph and $Q(1,-1)$ does not. $Q$ lands at $(1,3)$ , moving along the vertical line $x=1$ .

Full solution

The figure places $P$ one unit left and four up, at $(-1,4)$ , and $Q$ one right and one down, at $(1,-1)$ .
The substitutions give
$-1+2(4)=7$ $1+2(-1)=-1$
Thus only $P$ satisfies the equation.

For a vertical move from $Q$ , keep $x=1$ :
$1+2y=7$ $y=3$
The new point $(1,3)$ checks because $1+6=7$ .

The first coordinate stays $1$ throughout this movement, so $Q$ moves along the vertical line
$x=1$

Answer

$P(-1,4)$ lies on the graph and $Q(1,-1)$ does not. $Q$ lands at $(1,3)$ , moving along the vertical line $x=1$ .

Key idea

A vertical move changes the second coordinate while leaving the first fixed.
Problem 6: Mark complete
Problem 7 Faded equation

The figure shows a line whose equation was printed as $2x-3y=k$ . Read and report both axis crossings, recover $k$ , and check that the marked point $R$ satisfies the recovered equation.

A line whose equation was printed as $2x-3y=k$ , with its marked point $R$ .

Text description of this figure

A coordinate grid. The horizontal x-axis runs from negative two to eight and the vertical y-axis from negative six to two, with gridlines, tick marks and number labels at every whole number, equal unit lengths on both axes, arrowheads at both ends of each axis, and the origin labeled 0. A straight line with arrowheads at both ends rises from left to right across the whole grid. It crosses the y-axis exactly on a gridline four units below the origin, at zero, negative four, and crosses the x-axis exactly on a gridline six units right of the origin, at six, zero; neither crossing is marked or labeled. A dot on the line labeled R only sits three units right of the y-axis and two units below the x-axis, at three, negative two. No equation and no coordinates are printed.
1. Hint 1
  At an axis crossing, one of the two terms becomes zero.
2. Hint 2
  Use either crossing to recover the constant, then read both coordinates of $R$ .
Answer

x-intercept $(6,0)$ ; y-intercept $(0,-4)$ ; $k=12$ ; $R(3,-2)$ satisfies the equation.

Full solution

The crossings read from the grid are $(6,0)$ and $(0,-4)$ .
Using the first gives
$k=2(6)-3(0)$ $k=12$
The second gives $2(0)-3(-4)=12$ as well.

The marked point is $R(3,-2)$ .
Its check is
$2(3)-3(-2)=12$
Thus it satisfies $2x-3y=12$ .

Answer

x-intercept $(6,0)$ ; y-intercept $(0,-4)$ ; $k=12$ ; $R(3,-2)$ satisfies the equation.

Key idea

A visible crossing can recover a missing equation constant, and another point checks that recovery.
Problem 7: Mark complete
Problem 8 Table and graph

A table for $3y=x+2$ lists the points $(-2,0)$ , $(1,1)$ , $(4,2)$ , and $(7,3)$ . A student says that these four points are the entire graph. Is the statement correct? Give a solution missing from the table.
1. Hint 1
  A table can list selected solutions without listing every solution.
2. Hint 2
  Try an input between two of the listed inputs.
Answer

No; for example, $(0,\frac23)$ is a missing solution.

Full solution

The equation does not restrict $x$ to the four listed inputs.
At $x=0$ ,
$3y=2$ $y=\frac23$
So $(0,\frac23)$ is another solution.
There are infinitely many possible values of $x$ , each with a corresponding $y$ -value, and the graph contains all of those points.

Answer

No; for example, $(0,\frac23)$ is a missing solution.

Key idea

A finite table lists only some of the infinitely many points on a line's graph.
Problem 8: Mark complete
Problem 9 Two lists

List A contains $(-3,5)$ and $(1,1)$ . List B contains $(2,0)$ and $(6,-4)$ . Check that all four points satisfy $x+y=2$ . One student draws the line through List A and another draws the line through List B. Must they draw the same line? Explain.
1. Hint 1
  Ask what the equation says about the second coordinate for every first coordinate.
2. Hint 2
  Solve for $y$ and compare the two lists with the resulting rule.
Answer

Yes; all four points satisfy $x+y=2$ , and both students draw its graph, the line $y=2-x$ .

Full solution

The checks are
$-3+5=2$ $1+1=2$ $2+0=2$ $6+(-4)=2$
So all four points satisfy $x+y=2$ .

Solving the equation gives
$y=2-x$
This is a linear equation, so its graph is one straight line, and every listed point lies on it; each list contains two distinct points.

Two distinct points on a line fix that line, so drawing either pair produces the graph of $y=2-x$ .
The two lists are different samples from one graph, so the students draw the same line.

Answer

Yes; all four points satisfy $x+y=2$ , and both students draw its graph, the line $y=2-x$ .

Key idea

Any two distinct solutions of one linear equation locate the same line: its graph.
Problem 9: Mark complete
Problem 10 Moved point records

Every point on $x-2y=4$ is moved 6 units right with its height unchanged. A student claims that the moved points are exactly the graph of $x-2y=10$ . The word "exactly" makes two claims: that every moved point lies on $x-2y=10$ , and that every point of $x-2y=10$ is one of the moved points. Test each claim, and say whether the student's claim is correct.
1. Hint 1
  A move right adds to the first coordinate without changing the second.
2. Hint 2
  Compare the value of the expression before and after adding 6 to the first coordinate.
Answer

Yes, the claim is correct: every moved point lies on $x-2y=10$ , and every point of $x-2y=10$ is one of the moved points.

Full solution

Write an original point as $(a,b)$ , so $a-2b=4$ .
The moved point is $(a+6,b)$ .
Its value of $x-2y$ is
$a+6-2b=4+6$ $a+6-2b=10$
Thus every moved point lies on $x-2y=10$ .

Conversely, suppose $(c,d)$ satisfies $c-2d=10$ .
The point $(c-6,d)$ has
$(c-6)-2d=10-6$
which is $4$ , so $(c-6,d)$ lies on $x-2y=4$ .
Moving it 6 units right gives back $(c,d)$ , so every point of $x-2y=10$ is one of the moved points.

Both parts hold, so the moved points are exactly the graph of $x-2y=10$ .

Answer

Yes, the claim is correct: every moved point lies on $x-2y=10$ , and every point of $x-2y=10$ is one of the moved points.

Key idea

Checking a coordinate change in both directions identifies the entire resulting graph.
Problem 10: Mark complete

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