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Solving Systems by Elimination: 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 erased row

A system starts with $5x+2y=19$ . Twice this equation minus the second equation gives $7x=21$ . What was the second equation, in standard form?
1. Hint 1
  Write twice the first equation with both sides doubled.
2. Hint 2
  The erased left and right sides are what must be subtracted to leave the recorded row.
Answer

$3x+4y=17$ .

Full solution

Twice the first equation is
$10x+4y=38$
To leave $7x$ on the left, the subtracted left side must be $3x+4y$ .
To leave $21$ on the right, the subtracted right side must be $17$ .

Therefore the second equation was $3x+4y=17$ .
Checking the combination gives
$10x+4y-(3x+4y)=7x$
and $38-17=21$ , as required.

Answer

$3x+4y=17$ .

Key idea

A recorded equation combination determines an erased equation when the other equation and the combination are known.
Problem 1: Mark complete
Problem 2 Terms on both sides

Solve the system for $(x,y)$ and check the pair in both original equations.
$7x=3y+17$ $3y=1-2x$
1. Hint 1
  Columns only line up when every variable term sits on the same side of each equation.
2. Hint 2
  Rewrite each equation in the form $ax+by=c$ , then compare the two coefficients of $y$ .
Answer

$(2,-1)$ .

Full solution

Subtracting $3y$ from both sides of the first equation gives
$7x-3y=17$
Adding $2x$ to both sides of the second equation gives
$2x+3y=1$

The coefficients of $y$ are now opposites, so adding the two equations cancels $y$ .
$9x=18$ $x=2$

Back-substituting into $2x+3y=1$ gives $4+3y=1$ , so $3y=-3$ and $y=-1$ .

In the first original equation, $7(2)=14$ and $3(-1)+17=14$ , so it holds.
In the second, $3(-1)=-3$ and $1-2(2)=-3$ , so it holds too.

Answer

$(2,-1)$ .

Key idea

Putting both equations in standard form lines up the columns, so a matching pair of coefficients becomes visible.
Problem 2: Mark complete
Problem 3 The paired totals

Solve the system for $(x,y)$ .
$2(x+y)+3y=16$ $3(x+y)-2y=11$
1. Hint 1
  Match the coefficients of the whole group $x+y$ .
2. Hint 2
  Multiply the first equation by $3$ and the second by $2$ , then subtract.
Answer

$(3,2)$ .

Full solution

The scaled equations are
$6(x+y)+9y=48$ $6(x+y)-4y=22$
Subtracting gives
$13y=26$ $y=2$

The first original equation becomes $2(x+2)+6=16$ , so $2x=6$ and $x=3$ .
Checking gives $2(3+2)+3(2)=16$ and
$3(3+2)-2(2)=11.$

Answer

$(3,2)$ .

Key idea

Scaling can match coefficients of an entire shared group, after which elimination works as usual.
Problem 3: Mark complete
Problem 4 Two rows of boards

Boards of lengths $a$ cm and $b$ cm are laid end to end in two rows. Two boards of length $a$ and three of length $b$ make a row $24$ cm long, so $2a+3b=24$ . Four boards of length $a$ and one of length $b$ make a row $28$ cm long, so $4a+b=28$ . Find $a$ and $b$ , then find how the length of either row changes when one board of length $a$ is replaced by one board of length $b$ .
1. Hint 1
  Solving the two equations together gives both lengths, once one variable's coefficients are matched by scaling.
2. Hint 2
  Doubling the first equation gives it the same $a$ coefficient as the second, so subtracting then cancels $a$ .
3. Hint 3
  After finding both lengths, a replacement changes a row by the new length minus the old length.
Answer

$a=6$ cm and $b=4$ cm; either row becomes $2$ cm shorter.

Full solution

Double the first equation.
$4a+6b=48$
Subtract the second original equation.
$5b=20$ $b=4$

Back-substitute into $4a+b=28$ .
$4a+4=28$ $a=6$
The checks are $2(6)+3(4)=24$ and $4(6)+4=28$ .

Replacing one board of length $a$ by one of length $b$ changes a row by $4-6=-2$ cm, whichever row it is.
So either row becomes $2$ cm shorter: the first becomes $6+3(4)+4=22$ cm and the second $3(6)+4+4=26$ cm.

Answer

$a=6$ cm and $b=4$ cm; either row becomes $2$ cm shorter.

Key idea

After solving for component amounts, use their difference to evaluate a replacement.
Problem 4: Mark complete
Problem 5 The calibration pair

A calibration uses $\frac{x}{2}+\frac{y}{3}=4$ and $\frac{x}{3}-\frac{y}{2}=-\frac53$ . Find $(x,y)$ and verify the pair in the fractional equations.
1. Hint 1
  Clear the denominators in each whole equation first.
2. Hint 2
  Then match one pair of variable coefficients and eliminate.
Answer

$(4,6)$ .

Full solution

Multiply both original equations by $6$ .
$3x+2y=24$ $2x-3y=-10$
Multiply these equations by $3$ and $2$ , respectively.
$9x+6y=72$ $4x-6y=-20$

Adding yields $13x=52$ , so $x=4$ .
In $3x+2y=24$ , this gives $12+2y=24$ , so $y=6$ .

The fractional checks are
$\frac42+\frac63=4$ $\frac43-\frac62=-\frac53$
Both original equations hold.

Answer

$(4,6)$ .

Key idea

Clearing denominators preserves a system and can prepare its coefficients for elimination.
Problem 5: Mark complete
Problem 6 The adjustable coefficient

In the system $kx-4y=21$ and $3x-2y=8$ , the letter $k$ stands for a fixed number. Find the value of $k$ for which the system has no solution, and state how many solutions the system has for every other value of $k$ .
1. Hint 1
  A system has no solution exactly when elimination leaves a false statement with no variable in it.
2. Hint 2
  Double the second equation so its $y$ term matches the first, subtract, and collect the $x$ terms into one product.
3. Hint 3
  Dividing by an expression that contains $k$ is allowed only for the values of $k$ that keep it nonzero.
Answer

$k=6$ gives no solution; every other value of $k$ gives exactly one solution.

Full solution

Doubling the second equation gives
$6x-4y=16$
Subtracting it from the first equation cancels $y$ .
$kx-6x=21-16$ $(k-6)x=5$

If $k=6$ , the left side is $0x$ , so the equation says $0=5$ .
That is false for every pair, so no pair satisfies both equations, and the system has no solution.

If $k$ is any other number, then $k-6$ is not zero, so dividing by it is allowed.
$x=\frac{5}{k-6}$
That single value of $x$ then gives a single value of $y$ from $3x-2y=8$ .
That pair satisfies the first equation as well, because twice the second equation added to the combined equation gives back the first: $(k-6)x+2(3x-2y)$ simplifies to $kx-4y$ , and $5+2(8)=21$ .
So the system has exactly one solution.

For example, $k=7$ gives $x=5$ , and then $15-2y=8$ gives $y=\frac72$ .
The first equation checks: $7(5)=35$ , $4\left(\frac72\right)=14$ , and $35-14=21$ .

Answer

$k=6$ gives no solution; every other value of $k$ gives exactly one solution.

Key idea

When a coefficient is unknown, elimination shows which value makes both variables cancel, and a false leftover statement then means no solution.
Problem 6: Mark complete
Problem 7 The two formats

The same system is recorded in two formats. Format A is $y=11-3x$ and $2x+y=7$ . Format B is $3x+y=11$ and $2x+y=7$ . Give a short opening step suited to each format, and finish each route.
1. Hint 1
  One format already has a variable isolated; the other has matching coefficients.
2. Hint 2
  In format B, compare the coefficients of $y$ to choose between adding and subtracting; either format then leaves one equation in $x$ to solve and back-substitute.
Answer

A: replace $y$ by $11-3x$ . B: subtract one equation from the other. Both give $(4,-1)$ .

Full solution

For A, replacement gives
$2x+(11-3x)=7$ $11-x=7$ $x=4$
Thus
$y=11-3(4)=-1.$

For B, the coefficients of $y$ are equal, so subtracting the second equation from the first gives
$x=4$
Subtracting the first from the second instead gives $-x=-4$ , the same value.
Back-substitution into $2x+y=7$ gives $y=-1$ .

The results agree.
Checking the common pair gives $-1=11-3(4)$ , $2(4)-1=7$ , and $3(4)-1=11$ , so either format represents the same solution.

Answer

A: replace $y$ by $11-3x$ . B: subtract one equation from the other. Both give $(4,-1)$ .

Key idea

The form in which a system is written can make replacement or cancellation the shorter opening.
Problem 7: Mark complete
Problem 8 Both rows replaced

A system is $x+3y=13$ and $2x-y=5$ . Sol replaces both equations: one by their sum, $3x+2y=18$ , and the other by the first minus the second, $-x+4y=8$ . He claims the new system has exactly the same solutions as the original. Is he correct? Justify your answer, then solve the new system.
1. Hint 1
  Each new equation is a sum or difference of true equations, so every solution of the original system satisfies it.
2. Hint 2
  For the other direction, two quantities are fixed once their sum and their difference are both known; combine the new equations so an original left side reappears, then scale.
Answer

Yes. Both systems have the single solution $(4,3)$ .

Full solution

A pair that solves the original system makes both original equations true.
Adding or subtracting true equations gives a true equation, so the pair satisfies $3x+2y=18$ and $-x+4y=8$ .
No solution is lost.

Now take any pair that solves the new system.
Adding the two new equations gives
$2x+6y=26$
Halving both sides gives
$x+3y=13$
Subtracting the second new equation from the first gives
$4x-2y=10$
Halving both sides gives
$2x-y=5$

So every solution of the new system also satisfies both original equations, and no solution is invented.
Sol is correct.

To solve the new system, double $3x+2y=18$ to get $6x+4y=36$ , then subtract $-x+4y=8$ .
$7x=28$ $x=4$
Back-substituting gives $12+2y=18$ , so $y=3$ .
The pair $(4,3)$ checks in all four equations: $4+9=13$ , $8-3=5$ , $12+6=18$ and $-4+12=8$ .

Answer

Yes. Both systems have the single solution $(4,3)$ .

Key idea

Replacing equations by combinations keeps the solutions when the original equations can be rebuilt from the new ones.
Problem 8: Mark complete
Problem 9 The two zero rows

After valid elimination, system A becomes $x-2y=4$ together with $0=0$ . System B becomes $x-2y=4$ together with $0=3$ . Elena says both systems have no solution, because neither last row contains a variable. Is she correct? Describe the solutions of each system and give two examples when possible.
1. Hint 1
  Ask whether the last row accepts every pair or rejects every pair.
2. Hint 2
  If it accepts every pair, the retained equation still restricts the coordinates.
Answer

No. A: infinitely many solutions, every pair $(x,y)$ with $x-2y=4$ (equivalently $x=2y+4$ ), such as $(4,0)$ and $(6,1)$ ; any two such pairs are accepted. B: no solution.

Full solution

For A, $0=0$ imposes no additional restriction.
The retained equation gives $x=2y+4$ , so each choice of $y$ gives a matching $x$ , and every pair on the line $x-2y=4$ is a solution.
For example, $(4,0)$ and $(6,1)$ both satisfy $x-2y=4$ .

For B, $0=3$ is false for every pair.
Consequently no pair satisfies both rows.
The absence of variables does not make the two outcomes alike: one row is always true, and the other is never true.

So Elena's conclusion holds for B, but not because the row lacks a variable; for A it is wrong.

Answer

No. A: infinitely many solutions, every pair $(x,y)$ with $x-2y=4$ (equivalently $x=2y+4$ ), such as $(4,0)$ and $(6,1)$ ; any two such pairs are accepted. B: no solution.

Key idea

After cancellation, the truth of the remaining numerical statement determines whether the retained equation describes solutions or none exist.
Problem 9: Mark complete
Problem 10 The sum that kept both variables

For $4x+5y=28$ and $4x+2y=16$ , Ivo adds the equations and gets $8x+7y=44$ . He says this new equation must be false, because adding did not remove a variable. Is he correct? Explain what adding did and did not achieve, then solve the system.
1. Hint 1
  Can the sum of two true equations ever be false? Ask separately what adding was meant to achieve here.
2. Hint 2
  Compare the two coefficients of $x$ : are they opposites or equal, and which operation cancels a pair like that?
Answer

No: $8x+7y=44$ is true for every solution of the system, but it removes no variable. The solution is $(2,4)$ .

Full solution

A pair that solves the system makes both equations true, and adding true equations gives a true equation.
So $8x+7y=44$ holds for every solution, and Ivo is wrong to call it false.

Adding did not eliminate anything, though.
The coefficients of $x$ are $4$ and $4$ , equal rather than opposite, so adding doubled the $x$ term instead of canceling it, and the new equation still has two unknowns.

Because the coefficients of $x$ are equal, subtracting the second equation from the first cancels $x$ .
$3y=12$ $y=4$
Back-substituting into $4x+2y=16$ gives $4x+8=16$ , so $x=2$ .

The pair checks, since $4(2)+5(4)=28$ and
$4(2)+2(4)=16.$
It satisfies Ivo's equation too, since
$8(2)+7(4)=44.$

Answer

No: $8x+7y=44$ is true for every solution of the system, but it removes no variable. The solution is $(2,4)$ .

Key idea

Every sum of true equations is true, but only adding opposite coefficients or subtracting equal ones eliminates a variable.
Problem 10: Mark complete

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