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Solving Systems by Substitution: 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 The first move

Solve the system for $(x,y)$ .
$4x+5y=6$ $2y-x=5$
1. Hint 1
  Look for a variable whose coefficient is $1$ or $-1$ in either equation.
2. Hint 2
  Isolate $x$ in the second equation, then put that expression into the first.
Answer

$(-1,2)$ .

Full solution

In $2y-x=5$ the variable $x$ has coefficient $-1$ , so isolating it needs a sign flip and no division.
Adding $x$ to both sides and subtracting $5$ gives
$x=2y-5$

Substitute into the first equation.
$4(2y-5)+5y=6$ $8y-20+5y=6$ $13y=26$ $y=2$

Back-substitution gives $x=2(2)-5$ , so $x=-1$ .
Checking the original equations gives $4(-1)+5(2)=6$ and $2(2)-(-1)=5$ , both true.

Answer

$(-1,2)$ .

Key idea

Isolating a variable whose coefficient is $-1$ needs no division, so that first step brings in no fraction.
Problem 1: Mark complete
Problem 2 The repeated group

Solve the system for $(x,y)$ .
$y-2=3x$ $2(y-2)+x=28$
1. Hint 1
  The entire group $y-2$ already equals a simple expression.
2. Hint 2
  Replace that group in the second equation, then return to the first.
Answer

$(4,14)$ .

Full solution

Replace $y-2$ by $3x$ in the second equation.
$2(3x)+x=28$ $7x=28$ $x=4$

The first equation then gives
$y-2=12$ $y=14$

In the original equations, $14-2=3(4)$ and $2(14-2)+4=28$ , so both hold.

Answer

$(4,14)$ .

Key idea

An equal expression may replace a whole repeated group as well as a single variable.
Problem 2: Mark complete
Problem 3 The collected terms

Solve the system for $(x,y)$ .
$2x+y=8+x$ $y-1=3(x-1)$
1. Hint 1
  Rearrange an equation until one variable is alone.
2. Hint 2
  The first equation gives $y$ without introducing a fraction; put that expression in the second.
3. Hint 3
  The value of $x$ need not be a whole number; use its exact value when you back-substitute.
Answer

$\left(\frac52,\frac{11}{2}\right)$ , or $(2.5,5.5)$ .

Full solution

Subtracting $x$ from the first equation and isolating $y$ gives
$y=8-x$
Substitute into the second equation.
$8-x-1=3(x-1)$ $7-x=3x-3$ $10=4x$ $x=\frac52$

Back-substitution gives $y=8-\frac52$ , so
$y=\frac{11}{2}.$
In the first original equation, both sides equal $\frac{21}{2}$ , since $2\left(\frac52\right)+\frac{11}{2}=\frac{21}{2}$ and
$8+\frac52=\frac{21}{2}.$
In the second, both sides equal $\frac92$ , since $\frac{11}{2}-1=\frac92$ and
$3\left(\frac52-1\right)=\frac92.$

Answer

$\left(\frac52,\frac{11}{2}\right)$ , or $(2.5,5.5)$ .

Key idea

Simplifying an equation that has variables on both sides before isolating a variable keeps the substitution short, and the solution pair need not be whole numbers.
Problem 3: Mark complete
Problem 4 The requested crossing

Two lines are described by $x-y=2$ and $4x+by=14$ , where $b$ is a fixed but missing number. Their crossing is required to have $y=1$ . Find $b$ and the crossing pair, and check the pair in the completed equations.
1. Hint 1
  The requested $y$ value determines $x$ in the first equation.
2. Hint 2
  Use both coordinates in the second equation to recover its missing coefficient.
Answer

$b=2$ ; crossing $(3,1)$ .

Full solution

With $y=1$ , the first equation gives
$x-1=2$ $x=3$
Substituting the pair into the second equation gives
$4(3)+b(1)=14$ $b=2$

The completed system is $x-y=2$ and $4x+2y=14$ .
The pair $(3,1)$ gives $3-1=2$ and
$4(3)+2(1)=14.$
Replacing $x$ by $y+2$ in the second equation leaves $6y=6$ , so this is its only crossing.

Answer

$b=2$ ; crossing $(3,1)$ .

Key idea

A specified coordinate at a crossing can determine a missing coefficient in one of the two equations.
Problem 4: Mark complete
Problem 5 The two panels

The figure offers two candidate graphs for the system $y=2x-1$ and $x+y=5$ . Which panel shows the correct two lines? Find the exact crossing pair by substitution, then use it to justify your choice.

Two candidate graphs for the system $y=2x-1$ and $x+y=5$ .

Text description of this figure

Two square coordinate grids, one above the other. The top grid is headed Panel A and the bottom grid Panel B. In each grid the x-axis and the y-axis run from negative 1 to 6 with equal scales, every whole number is labeled, and the origin is labeled 0. Each grid shows two solid straight lines with arrowheads where they leave the grid, and no equations, point labels or dots. In both panels one line, drawn in the text color, falls from the top-left corner (negative 1, 6) through (0, 5) and (5, 0) to the bottom-right corner (6, negative 1). In Panel A the other line, drawn in blue, rises 2 units for every 1 unit to the right: it leaves the bottom edge at (0, negative 1), crosses the x-axis halfway between 0 and 1, and leaves the top edge halfway between x equals 3 and x equals 4. In Panel B the blue line also rises 2 units for every 1 unit to the right: it leaves the bottom edge halfway between x equals 1 and x equals 2, crosses the x-axis at 2, and leaves the top edge at (5, 6). In each panel the two lines cross once inside the grid, and the crossing is not marked or labeled.
1. Hint 1
  Find the common pair before comparing the panels.
2. Hint 2
  Put the expression for $y$ into $x+y=5$ .
3. Hint 3
  In each panel, read a crossing as $x$ across and $y$ up; swapping the two coordinates names a different point.
Answer

Panel A; $(2,3)$ .

Full solution

Substitution gives
$x+(2x-1)=5$ $3x=6$ $x=2$
Then $y=2(2)-1$ , so $y=3$ .

The pair checks in both original equations: $3=2(2)-1$ and $2+3=5$ .

Panel B's lines cross at $(3,2)$ , which fails $y=2x-1$ because $2(3)-1$ is $5$ , not $2$ .
So Panel B cannot show this system.
Panel A's lines cross at $(2,3)$ .
Its rising line passes through $(0,-1)$ and $(2,3)$ , which both satisfy $y=2x-1$ , and its falling line passes through $(0,5)$ and $(5,0)$ , which both satisfy $x+y=5$ .
Therefore Panel A is correct.

Answer

Panel A; $(2,3)$ .

Key idea

The common ordered pair found algebraically identifies the crossing in a graph.
Problem 5: Mark complete
Problem 6 Mina's rewritten rule

A pair $(x,y)$ satisfies both rules $x=(y+7)/3$ and $2x-3y=-7$ . Mina first rewrites the first rule as $3x-y=7$ . Continue from that form using an isolation that introduces no fractions, find the pair, and check it in the two original rules.
1. Hint 1
  In Mina's form, the coefficient of $y$ is $-1$ .
2. Hint 2
  Isolate $y$ in Mina's form, then put that expression into $2x-3y=-7$ inside parentheses.
Answer

$(4,5)$ .

Full solution

Isolating $x$ or $y$ in $2x-3y=-7$ would mean dividing by $2$ or $3$ , and isolating $x$ in Mina's form would mean dividing by $3$ .
Only $y$ in Mina's form has coefficient $-1$ , so isolate it.
$y=3x-7$

Replace $y$ in the second rule, keeping the parentheses so the minus sign reaches both terms.
$2x-3(3x-7)=-7$ $2x-9x+21=-7$ $-7x=-28$ $x=4$

Back-substitution gives
$y=3(4)-7$ $y=5$
Checking the original rules gives $4=(5+7)/3$ and $2(4)-3(5)=-7$ , both true.

Answer

$(4,5)$ .

Key idea

Choosing a variable with coefficient one or negative one can avoid fractions even when a given rule contains a fraction.
Problem 6: Mark complete
Problem 7 Two closing reports

The first equation in each system is $y=3x+2$ . In system A, the second equation is $2(y-1)=6x+2$ . In system B, it is $2(y-1)=6x+8$ . Decide how many solutions each system has and describe how its two lines sit.
1. Hint 1
  Use the same expression for $y$ in each second equation.
2. Hint 2
  An identity keeps every pair on the first line; a false numerical statement keeps none.
Answer

A: infinitely many, the same line. B: no solution, distinct parallel lines.

Full solution

For A, replacement gives
$2(3x+2-1)=6x+2$ $6x+2=6x+2$
This is true for every $x$ , so A has all pairs $(x,3x+2)$ and its equations describe the same line.

For B, replacement gives
$2(3x+2-1)=6x+8$ $6x+2=6x+8$ $2=8$
This is impossible.
The second equation can also be written $y=3x+5$ , so B has two distinct parallel lines.

Answer

A: infinitely many, the same line. B: no solution, distinct parallel lines.

Key idea

After substitution cancels the variable, the remaining constants distinguish one repeated line from two parallel lines.
Problem 7: Mark complete
Problem 8 Teo's repair

The pair $(3,2)$ fits $y=5-x$ but fails $2x-y=1$ . Teo keeps $y=2$ and changes only $x$ until the second equation holds. He claims this repairs the solution of the system. What pair does he obtain, and is his claim correct?
1. Hint 1
  Decide what a pair must satisfy to solve the whole system.
2. Hint 2
  With $y=2$ , solve the second equation for $x$ , then test the changed pair in the first.
Answer

$\left(\frac32,2\right)$ , or $(1.5,2)$ ; the claim is incorrect.

Full solution

Keeping $y=2$ in the second equation gives
$2x-2=1$ $2x=3$ $x=\frac32$
Thus Teo obtains $\left(\frac32,2\right)$ .

This pair satisfies the second equation, but the first would require
$y=5-\frac32$ $y=\frac72$
Its actual second coordinate is $2$ , so the pair fails the first equation.
Fixing the second condition alone has disturbed the first, and the system is not solved.

Answer

$\left(\frac32,2\right)$ , or $(1.5,2)$ ; the claim is incorrect.

Key idea

A repaired coordinate must be tested in every original condition, including one the earlier pair satisfied.
Problem 8: Mark complete
Problem 9 The second route

For $x+2y=9$ and $y=4-x$ , Arun replaces $y$ in the first equation. Bea first rewrites the second equation as $x=4-y$ and replaces $x$ in the first equation. Bea says both routes must lead to the same pair. Is she correct? Carry out both routes, then explain why two valid routes could not give different pairs.
1. Hint 1
  Each expression used for replacement comes from the same second equation.
2. Hint 2
  One route leaves only $x$ , while the other leaves only $y$ .
Answer

Yes; both routes give $(-1,5)$ .

Full solution

Arun obtains
$x+2(4-x)=9$ $8-x=9$ $x=-1$
The second equation then gives $y=5$ .

Bea obtains
$(4-y)+2y=9$ $4+y=9$ $y=5$
Her expression for $x$ then gives $x=-1$ .

Each route replaces a quantity by an equal one, so its one-variable equation, together with the second equation, is satisfied by exactly the pairs that solve the system.
Both routes therefore find the solutions of the same system, so two valid routes could not give different pairs.
The pair checks because $-1+2(5)=9$ and $5=4-(-1)$ , so Bea is correct.

Answer

Yes; both routes give $(-1,5)$ .

Key idea

Valid isolations of the same equation can produce different one-variable routes to the same common pair.
Problem 9: Mark complete
Problem 10 A shifted system

A system has equations $y=4x-7$ and $x+2y=4$ . In a new system, they become $y=4x-5$ and $x+2y=8$ . Noor says every solution of the old system becomes a solution of the new one by keeping $x$ and adding $2$ to $y$ . Is this correct, and can the change be reversed? Explain without solving either system.
1. Hint 1
  Write a new pair as $(x,y+2)$ using an old solution $(x,y)$ .
2. Hint 2
  Check each new equation using the corresponding old equation, then consider subtracting $2$ .
Answer

Yes; adding $2$ to $y$ works, and subtracting $2$ reverses it.

Full solution

For an old solution, $y=4x-7$ , so
$y+2=4x-5$
Thus the changed pair satisfies the first new equation.

The old pair also has $x+2y=4$ .
Adding $4$ to both sides gives $x+2y+4=8$ , which is
$x+2(y+2)=8$
so the changed pair satisfies the second new equation.

Conversely, if $(x,y)$ solves the new system, then $y-2=4x-7$ and $x+2(y-2)=4$ , so subtracting $2$ from the second coordinate gives an old solution.
The change therefore matches the solution pairs in both directions.

Answer

Yes; adding $2$ to $y$ works, and subtracting $2$ reverses it.

Key idea

Check a proposed change of coordinates in every equation and in reverse to establish that it matches two solution sets.
Problem 10: Mark complete

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