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Equations with Two Variables: 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 Variables on both sides

Write $2(x+3y)=5(x-1)+y$ in the form $ax+by=c$ , with integer coefficients having no common factor greater than $1$ and with $a$ positive.
1. Hint 1
  The target form has every variable term on the left and a single number on the right; notice that both $x$ and $y$ appear on both sides.
2. Hint 2
  Expand both groups first, then move every variable term to one side and every constant to the other.
3. Hint 3
  If the coefficient of $x$ comes out negative, multiply both sides by $-1$ .
Answer

$3x-5y=5$ .

Full solution

Expanding both groups gives
$2x+6y=5x-5+y$
Subtract $5x$ and $y$ from both sides.
$-3x+5y=-5$
Multiply both sides by $-1$ so that the coefficient of $x$ is positive.
$3x-5y=5$

The coefficients $3$ , $-5$ and $5$ are integers with no common factor greater than $1$ .
As a check, $(0,-1)$ satisfies $3x-5y=5$ , since
$3(0)-5(-1)=5.$
In the original equation it makes both sides $-6$ , since $2(0-3)=-6$ and
$5(0-1)+(-1)=-6.$

Answer

$3x-5y=5$ .

Key idea

Expanding and then collecting like terms turns an equation with variables on both sides into standard form.
Problem 1: Mark complete
Problem 2 The shifted reading

A reading is stored as $(x,y)$ . The instrument adds $1$ to its first coordinate and leaves its second coordinate unchanged. Which original reading, $(1,5)$ or $(2,8)$ , produces a new pair satisfying $4x-y=3$ ?
1. Hint 1
  Test the pair after the instrument changes it.
2. Hint 2
  The new first coordinate belongs in the $x$ position.
Answer

$(1,5)$ .

Full solution

The original reading $(1,5)$ becomes $(2,5)$ .
$4(2)-5=3$
It satisfies the equation.

The original reading $(2,8)$ becomes $(3,8)$ .
$4(3)-8=4$
It does not satisfy the equation.

Answer

$(1,5)$ .

Key idea

Test the coordinates of the transformed pair in their stated order.
Problem 2: Mark complete
Problem 3 A pair to choose

Give one solution of $2x-3y=-12$ whose first coordinate is negative and whose second coordinate is positive.
1. Hint 1
  Choose a negative value for the first coordinate, then solve for the second.
2. Hint 2
  Try a small negative integer, then check the sign of the resulting $y$ .
Answer

For example, $(-3,2)$ .

Full solution

Choose $x=-3$ and substitute.
$-6-3y=-12$ $-3y=-6$ $y=2$
Thus $(-3,2)$ is one valid answer.

Checking, $2(-3)-3(2)=-12$ , and its coordinates have the required signs.
Any other pair with a negative first coordinate and a positive second coordinate that satisfies the equation is also accepted.

Answer

For example, $(-3,2)$ .

Key idea

A chosen first coordinate produces a solution when the matching second coordinate is found from the equation.
Problem 3: Mark complete
Problem 4 The paper strip

A maker starts with a red strip of length $r$ cm and a blue strip of length $b$ cm. Three copies of the red strip are each extended by $2$ cm, and four copies of the blue strip are each shortened by $1$ cm. The seven finished strips total $40$ cm. Write an equation relating $r$ and $b$ in standard form, with the $r$ term first and integer coefficients having no common factor greater than $1$ , then give two possible original pairs $(r,b)$ with $r>0$ and $b>1$ .
1. Hint 1
  Each copy contributes its finished length to the total.
2. Hint 2
  Expand $3(r+2)+4(b-1)$ , then choose convenient values for one length.
Answer

$3r+4b=38$ ; for example, $(10,2)$ and $(2,8)$ , in cm.

Full solution

The finished strips give
$3(r+2)+4(b-1)=40$ $3r+6+4b-4=40$ $3r+4b=38$

Choosing $b=2$ gives $3r=30$ , so
$(r,b)=(10,2).$
Choosing $b=8$ gives $3r=6$ , so
$(r,b)=(2,8).$

Both pairs satisfy the length restrictions.
The finished totals are $3(12)+4(1)=40$ cm and $3(4)+4(7)=40$ cm.
Any two distinct pairs with $r>0$ and $b>1$ satisfying the equation are accepted.

Answer

$3r+4b=38$ ; for example, $(10,2)$ and $(2,8)$ , in cm.

Key idea

Write the amount contributed by each changed item before collecting a total equation.
Problem 4: Mark complete
Problem 5 The two blank marks

The line drawn is the graph of the equation $x-y=-3$ . Check that both marked points satisfy the equation. Then use the equation to find the solution whose first coordinate is $10$ , which lies beyond the grid, and explain why it is on the graph even though it is not drawn.

The graph of one linear equation, with two of its solutions marked.

Text description of this figure

A square coordinate grid. The horizontal x-axis and the vertical y-axis each run from negative four to six, with tick marks, number labels and gridlines at every whole number and equal unit lengths on both axes, and the origin labeled 0. One straight line with an arrowhead at each end rises from the lower left to the upper right across the whole grid. It enters at the left edge of the grid, at the point (negative 4, negative 1), and leaves at the top edge, at the point (3, 6). Two points on the line are marked with solid dots and carry no labels: one on the x-axis at (negative 3, 0), and one at (1, 4). No equation, coordinate labels or guide lines are shown.
1. Hint 1
  Read each point across first, then up, and write it as an ordered pair.
2. Hint 2
  Substitute the first coordinate for $x$ and the second for $y$ in $x-y$ , and compare the result with $-3$ .
3. Hint 3
  Put $x=10$ into $x-y=-3$ and solve for $y$ ; then recall what the graph of an equation contains.
Answer

$(-3,0)$ and $(1,4)$ both satisfy the equation; the solution with first coordinate $10$ is $(10,13)$ , and it lies on the graph.

Full solution

The marked points are $(-3,0)$ and $(1,4)$ .
Substituting the first gives
$-3-0=-3$
Substituting the second gives
$1-4=-3$
Both pairs make the equation true.

Put $x=10$ into the equation.
$10-y=-3$
Subtract $10$ from both sides.
$-y=-13$ $y=13$
The solution is $(10,13)$ , and checking gives $10-13=-3$ .

The graph of an equation is the set of all its solutions, so every solution is a point on the line, whether or not it fits in the picture.
The arrowheads show that the line continues past the grid, and $(10,13)$ is one of its points out there.

Answer

$(-3,0)$ and $(1,4)$ both satisfy the equation; the solution with first coordinate $10$ is $(10,13)$ , and it lies on the graph.

Key idea

The graph of an equation contains every one of its solutions, including those beyond the part that is drawn.
Problem 5: Mark complete
Problem 6 An added condition

A pair $(x,y)$ must satisfy $3x+y=11$ . A recorder can add either condition A, $x=2$ , or condition B, $6x+2y=22$ . Which condition fixes exactly one pair, and what is that pair? Explain what the other condition contributes.
1. Hint 1
  Ask whether each condition adds information to the first equation.
2. Hint 2
  One condition gives a coordinate directly; for the other, ask whether it can be obtained from $3x+y=11$ by doing the same thing to both sides.
Answer

Condition A fixes $(2,5)$ . Condition B adds no new restriction.

Full solution

With condition A, substituting $x=2$ gives
$3(2)+y=11$ $y=5$
Thus A fixes $(2,5)$ , and $3(2)+5=11$ checks it.
On the graph, the line $x=2$ crosses the line $3x+y=11$ at that one point.

Condition B is twice the original equation.
Every pair satisfying $3x+y=11$ already satisfies $6x+2y=22$ , so B leaves the same whole line of possible pairs.

Answer

Condition A fixes $(2,5)$ . Condition B adds no new restriction.

Key idea

A second equation that is a multiple of the first adds no restriction, while one whose line crosses the first line once fixes a single pair.
Problem 6: Mark complete
Problem 7 Describing every solution

Find every solution pair of $4x+2y=4x+10$ , where $x$ and $y$ may be any real numbers. Give two of them, and describe the full graph.
1. Hint 1
  The equal $x$ terms can be removed from both sides.
2. Hint 2
  After finding $y$ , decide whether the equation places any restriction on $x$ .
Answer

Every pair $(x,5)$ , for any real $x$ ; for example, $(0,5)$ and $(2,5)$ ; the graph is the horizontal line $y=5$ .

Full solution

Subtract $4x$ from both sides.
$2y=10$ $y=5$
No condition on $x$ remains.

Thus every pair $(x,5)$ is a solution.
For example, $(0,5)$ gives $10=10$ in the original equation, and $(2,5)$ gives $18=18$ .
All the points sit $5$ units above the horizontal axis, forming the horizontal line $y=5$ .

Answer

Every pair $(x,5)$ , for any real $x$ ; for example, $(0,5)$ and $(2,5)$ ; the graph is the horizontal line $y=5$ .

Key idea

If one variable cancels from an equation, its coordinate may remain free while the other coordinate is fixed.
Problem 7: Mark complete
Problem 8 Rae's one solution

Rae says, "The equation $5x+3y=22$ has just one solution, $(2,4)$ ." Is this correct when $x$ and $y$ may be any real numbers? Is it correct when both must be positive whole numbers? Explain both decisions.
1. Hint 1
  The kind of numbers allowed changes which pairs count as solutions.
2. Hint 2
  For real numbers, choose a different value of $x$ and solve for the matching $y$ .
3. Hint 3
  For positive whole numbers, $y$ is at least $1$ , so $5x$ is at most $19$ ; test each whole number value of $x$ that this allows.
Answer

Any real numbers: no; for example, $(5,-1)$ also works. Positive whole numbers: yes, $(2,4)$ is the only one.

Full solution

For real numbers, Rae is incorrect.
The pair $(5,-1)$ is another solution, since
$5(5)+3(-1)=22$
In fact, any chosen value of $x$ gives a matching $y$ , so there are infinitely many solutions.

For positive whole numbers, $y$ is at least $1$ , so $3y$ is at least $3$ and $5x$ is at most $19$ .
So $x$ can only be $1$ , $2$ or $3$ .

If $x=1$ , then $3y=17$ , and $17$ is not a multiple of $3$ , so $y$ is not a whole number.
If $x=2$ , then $3y=12$ , so $y=4$ .
If $x=3$ , then $3y=7$ , and $7$ is not a multiple of $3$ .

So $(2,4)$ is the only solution in positive whole numbers, and Rae is correct under that restriction.

Answer

Any real numbers: no; for example, $(5,-1)$ also works. Positive whole numbers: yes, $(2,4)$ is the only one.

Key idea

Distinguish the full solution set of an equation from the pairs left by a separate restriction on the allowed numbers.
Problem 8: Mark complete
Problem 9 Jo's new pairs

A pair $(x,y)$ satisfies $3x-y=4$ . Jo makes a new pair $(x+t,y+mt)$ by adding a real number $t$ to the first coordinate and $mt$ to the second, where $m$ is a fixed number. For which value of $m$ does the new pair also satisfy the equation for every real $t$ ? Explain.
1. Hint 1
  The new pair is a solution exactly when the change it makes to $3x-y$ adds up to nothing.
2. Hint 2
  Substitute the new pair into the left side of the equation, remembering that $3x-y$ is already known, then group the terms that contain $t$ apart from the rest.
3. Hint 3
  The left side must equal $4$ whatever $t$ is; try $t=1$ to see what the terms containing $t$ must add to.
Answer

$m=3$ .

Full solution

Substitute the new pair into the left side.
$3(x+t)-(y+mt)$
Expanding and grouping the terms that contain $t$ gives
$3x-y+3t-mt$
Since $3x-y=4$ and $3t-mt=(3-m)t$ , the left side equals
$4+(3-m)t$

If $m=3$ , then $(3-m)t=0$ for every $t$ , so the left side is $4$ and the new pair is always a solution.

If $m$ is any other number, choose $t=1$ .
The left side is then $4+(3-m)$ , which is not $4$ because $3-m$ is not $0$ , so the new pair fails.

Therefore $m=3$ .
For example, $(1,-1)$ satisfies the equation, and $t=3$ with $m=3$ gives $(4,8)$ , where $3(4)-8=4$ .

Answer

$m=3$ .

Key idea

From one solution of $3x-y=4$ , a step of $t$ in $x$ keeps the equation true only with a step of $3t$ in $y$ , which is why the solutions line up.
Problem 9: Mark complete
Problem 10 Liv's verification

A system consists of $x+2y=8$ and $3x-y=6$ . Liv verifies $(2,3)$ in the first equation and $(2,0)$ in the second, then says the system is solved. Is this sufficient? Explain, and decide whether either checked pair solves the system.
1. Hint 1
  A system requires the same pair to satisfy every equation.
2. Hint 2
  Test each checked pair in the equation Liv did not use for it.
Answer

No; neither $(2,3)$ nor $(2,0)$ solves the system.

Full solution

The first pair does satisfy the first equation, since $2+2(3)=8$ .
But in the second equation it gives
$3(2)-3=3$
This is not $6$ , so the first pair fails the second equation.

The second pair satisfies $3(2)-0=6$ , but gives $2+2(0)=2$ in the first equation, not $8$ .

Liv's checks are not sufficient: a solution of the system is one pair that satisfies both equations, and neither checked pair does.

Answer

No; neither $(2,3)$ nor $(2,0)$ solves the system.

Key idea

A system is solved by a common pair, rather than separate pairs chosen for its separate equations.
Problem 10: Mark complete

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