Core practice ← Back to lesson

Solving 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)

0 of 10 completed

Progress saved in this browser.

Problem 1 of 10

Problem 1 Two operations around one variable

Solve $4x - 7 = -17$ , and check your value in the original equation.
1. Hint 1
  Read $4x - 7$ as a recipe carried out on $x$ : one operation was applied first and one last. Solving runs the recipe backwards.
2. Hint 2
  The last operation applied was subtracting $7$ , so undo it first by adding $7$ to both sides. That leaves $4x$ alone on the left.
3. Hint 3
  One more move, applied to both sides, undoes the multiplication by $4$ . The value need not be a whole number; check it in $4x - 7$ itself.
Answer

$x = -\tfrac{5}{2}$ (or $-2.5$ ); check: $4(-\tfrac52) - 7 = -17$ .

Full solution

Building $4x - 7$ from $x$ multiplies by $4$ and then subtracts $7$ .
Undoing them in reverse order, the subtraction first, is the convenient order, because it leaves $4x$ on its own.
It is not the only legal order.

Add $7$ to both sides:
$4x = -10.$
Divide both sides by $4$ , which is allowed because $4$ is not zero:
$x = -\frac{10}{4}.$
Simplify the fraction by dividing its top and bottom by $2$ :
$x = -\frac{5}{2}.$
As a decimal, this is $-2.5$ .

Check in the original equation: $4\left(-\tfrac{5}{2}\right) = -10$ , and $-10 - 7 = -17$ , which matches the right side.

Dividing by $4$ first is legal too, but the division must reach every term, the $7$ included.
It gives $x - \tfrac{7}{4} = -\tfrac{17}{4}$ , and adding $\tfrac{7}{4}$ to both sides then gives the same $x = -\tfrac{5}{2}$ , with more fractions along the way.

Answer

$x = -\tfrac{5}{2}$ (or $-2.5$ ); check: $4(-\tfrac52) - 7 = -17$ .

Key idea

Undo the operations around the variable in the reverse of the order they were applied, and let a property of equality license each move.
Problem 1: Mark complete
Problem 2 A coefficient that is a fraction

Solve $\dfrac{3}{5}x - 4 = 8$ , and check your value in the original equation.
1. Hint 1
  A fractional coefficient is handled like a whole-number one: remove the constant first, then undo the multiplication with its inverse.
2. Hint 2
  After adding $4$ to both sides, a fraction of $x$ equals a number. Which number, multiplied by $\tfrac{3}{5}$ , gives exactly $1$ ?
Answer

$x = 20$ ; check: $\tfrac35(20) - 4 = 8$ .

Full solution

Undo the operations in reverse order.
The constant was attached last, so remove it first by adding $4$ to both sides:
$\frac{3}{5}x = 12.$

Now $x$ is multiplied by $\tfrac{3}{5}$ .
That coefficient is not zero, so it has a reciprocal, $\tfrac{5}{3}$ ; zero has no reciprocal, which is why this move needs a nonzero coefficient.
Multiply both sides by $\tfrac{5}{3}$ .
On the left, $\tfrac{5}{3} \cdot \tfrac{3}{5} = 1$ , which leaves $1 \cdot x$ , that is, $x$ .
On the right,
$\tfrac{5}{3} \cdot 12 = 20.$
So $x = 20$ .

Check in the original equation: $\tfrac{3}{5}(20) = 12$ , and $12 - 4 = 8$ , which matches the right side.

Multiplying both sides by $\tfrac{5}{3}$ is the multiplication property of equality, allowed because $\tfrac{5}{3}$ is not zero.
Dividing both sides by $\tfrac{3}{5}$ , the division property of equality, is the same move under another name, because dividing by a nonzero number means multiplying by its reciprocal.

Answer

$x = 20$ ; check: $\tfrac35(20) - 4 = 8$ .

Key idea

Multiplying both sides by the reciprocal of a nonzero fractional coefficient turns that coefficient into $1$ and leaves the variable alone.
Problem 2: Mark complete
Problem 3 An equation with parentheses

Solve $9x - 4(x + 3) = 18$ , and check your value in the original equation.
1. Hint 1
  Simplify the left side completely before moving anything across the equals sign. A minus sign in front of a group belongs to every term inside it.
2. Hint 2
  Expand $4(x + 3)$ to $4x + 12$ , keep it in parentheses, and subtract that whole group from $9x$ .
Answer

$x = 6$ ; the check gives $9(6) - 4(9) = 54 - 36 = 18$ .

Full solution

Simplify the left side on its own before moving anything.
The $-4$ multiplies both terms in the parentheses, so the $3$ contributes $-12$ :
$9x - 4(x + 3) = 9x - 4x - 12.$

Combine the like terms $9x$ and $-4x$ to get $5x$ , so the equation is
$5x - 12 = 18.$
Add $12$ to both sides:
$5x = 30.$
Divide both sides by $5$ :
$x = 6.$

Check in the original equation, working out the parentheses first: $9(6) - 4(6 + 3) = 54 - 36 = 18$ , which matches the right side.

Answer

$x = 6$ ; the check gives $9(6) - 4(9) = 54 - 36 = 18$ .

Key idea

When a group is subtracted, the negative factor multiplies every term inside the parentheses, so distribute it to every term, not just the first.
Problem 3: Mark complete
Problem 4 One equation, collected two ways

Solve $3x + 14 = 8x - 6$ twice: once by first subtracting $3x$ from both sides, and once by first subtracting $8x$ from both sides. Write the equation after each move, and check your value in the original equation.

Which route would you recommend to someone who often makes sign slips, and which line of your work shows why?
1. Hint 1
  Both routes use legal moves. The real question is what each one costs you along the way.
2. Hint 2
  Subtracting $8x$ leaves the variable with a negative coefficient. Keep that minus sign attached in every line, including the final division.
Answer

Both routes give $x = 4$ ; check: both sides of the original equal $26$ . Recommend subtracting $3x$ first; the line that shows why is $14 = 5x - 6$ .

Full solution

On the first route, subtract $3x$ from both sides.
The variable terms cancel on the left, and on the right $8x - 3x = 5x$ :
$14 = 5x - 6.$
Add $6$ to both sides:
$20 = 5x.$
Divide both sides by $5$ :
$x = 4.$

On the second route, subtract $8x$ from both sides.
Now the variable terms cancel on the right, and on the left $3x - 8x = -5x$ :
$-5x + 14 = -6.$
Subtract $14$ from both sides:
$-5x = -20.$
Divide both sides by $-5$ ; a negative divided by a negative is positive:
$x = 4.$

Check in the original equation: the left side is $3(4) + 14 = 26$ and the right side is
$8(4) - 6 = 26.$
The two agree.

Recommend the first route to someone who slips on signs.
Its line $14 = 5x - 6$ has a positive coefficient, and no negative number appears after it.
The second route's line $-5x + 14 = -6$ carries a negative coefficient and ends with a division by $-5$ , two extra chances to lose a sign.

Answer

Both routes give $x = 4$ ; check: both sides of the original equal $26$ . Recommend subtracting $3x$ first; the line that shows why is $14 = 5x - 6$ .

Key idea

Both routes are legal, but subtracting the smaller variable term keeps the coefficient positive and leaves fewer signs to get wrong.
Problem 4: Mark complete
Problem 5 Keeping a balance level

A balance scale is level. Its left pan holds four identical bags of flour and a $2$ kg weight, and its right pan holds a single bag of the same kind and an $11$ kg weight. Writing $b$ for the weight of one bag in kilograms, the level scale says
$4b + 2 = b + 11.$
Solve for $b$ , describing each move as something done to the two pans, and check your value by finding what each pan weighs. Then explain why a move made with the bags can keep the scale level even though nobody yet knows what a bag weighs.
1. Hint 1
  Removing equal weights from the two pans keeps the scale level.
2. Hint 2
  Lift a bag off each pan so that bags remain on the left only. Then take the loose weight off both pans, and share out what is left.
3. Hint 3
  A letter stands for one definite number, even while you do not know it. Ask whether both pans lose the same weight when a bag comes off each.
Answer

$b = 3$ : each bag weighs $3$ kg; check: each pan weighs $14$ kg. Moves: a bag off each pan, $2$ kg off each pan, then a third of each pan kept.

Full solution

Collect the bags on the left, where there are more of them.
Subtracting $b$ from both sides is lifting one bag off each pan:
$4b + 2 - b = b + 11 - b.$
Combining like terms on each side leaves
$3b + 2 = 11.$

Subtracting $2$ from both sides is taking $2$ kg off each pan:
$3b = 9.$
Dividing both sides by $3$ is keeping a third of what is on each pan, one of the three bags on the left and $3$ kg of the $9$ kg on the right.
That leaves $b = 3$ .

Check against the scale as described: the left pan holds $4(3) + 2 = 14$ kg and the right pan holds $3 + 11 = 14$ kg.
The pans balance, so each bag weighs $3$ kg.

The weight of a bag is unknown to us, but it is not undecided: every bag weighs the same definite number of kilograms, and $b$ names that number.
The pans hold equal weights before the move.
Lifting one bag off each takes the same weight, $b$ kg, from both, and equal amounts with the same amount removed stay equal.
That is the subtraction property of equality, used with the number $b$ .

Answer

$b = 3$ : each bag weighs $3$ kg; check: each pan weighs $14$ kg. Moves: a bag off each pan, $2$ kg off each pan, then a third of each pan kept.

Key idea

Removing equal weights from both pans, like subtracting the same number from both sides, keeps the balance level even when the weight of a bag is unknown.
Problem 5: Mark complete
Problem 6 Once for the letter, once for a missing number

Solve $4(x - 3) + 2x = 3(x + 1)$ , and check your value in the original equation.

Then find the number $k$ for which the equation $4(x - 3) + 2x = 3(x + k)$ has the solution $x = 11$ , and check that your value of $k$ works.
1. Hint 1
  Tidy each side on its own until it is as simple as it can be, then collect the variable on one side and strip away what surrounds it.
2. Hint 2
  Distribute on each side separately and combine the two $x$ terms on the left. Then collect the variable on the side with the larger coefficient.
3. Hint 3
  For $k$ , the value of $x$ is given, so substitute $x = 11$ before anything else. The left side becomes a number, and $k$ is the only letter left.
Answer

$x = 5$ , where both sides equal $18$ . The missing number is $k = 7$ : at $x = 11$ the left side is $54$ , and $3(11 + 7) = 54$ .

Full solution

Simplify each side on its own before moving anything across the equals sign.
The right side distributes to $3x + 3$ .
On the left, distribute and then combine the variable terms:
$\begin{aligned} 4(x - 3) + 2x &= 4x - 12 + 2x \\ &= 6x - 12. \end{aligned}$

The equation now reads
$6x - 12 = 3x + 3.$
Subtract $3x$ from both sides, which keeps the variable on the left with a positive coefficient:
$3x - 12 = 3.$
Add $12$ to both sides:
$3x = 15.$
Divide both sides by $3$ :
$x = 5.$

Check in the original equation: the left side is $4(5 - 3) + 2(5) = 8 + 10 = 18$ and the right side is
$3(5 + 1) = 18.$
The two agree.

For the second task the value of $x$ is known and $k$ is not, so substitute $x = 11$ first.
The left side then contains no unknown at all:
$\begin{aligned} 4(11 - 3) + 2(11) &= 32 + 22 \\ &= 54. \end{aligned}$

The right side becomes $3(11 + k)$ , so what has to hold is
$3(11 + k) = 54.$

Isolate $k$ with the same moves that isolate $x$ : distribute, subtract $33$ from both sides, then divide both sides by $3$ :
$\begin{aligned} 33 + 3k &= 54 \\ 3k &= 21 \\ k &= 7. \end{aligned}$

Check: with $k = 7$ , the right side at $x = 11$ is $3(11 + 7) = 3(18) = 54$ , which matches the left side.
Solving $4(x - 3) + 2x = 3(x + 7)$ directly confirms it, since $6x - 12 = 3x + 21$ gives $3x = 33$ and $x = 11$ .

Answer

$x = 5$ , where both sides equal $18$ . The missing number is $k = 7$ : at $x = 11$ the left side is $54$ , and $3(11 + 7) = 54$ .

Key idea

The properties of equality do not care which letter is unknown: once the other letter has a value, the same moves isolate the one that is left.
Problem 6: Mark complete
Problem 7 Two transit cards running down

Ana and Ben start the week with money on their transit cards, and each rides every day. For as long as both cards still hold money, after $t$ days Ana's card holds $45 - 5t$ dollars and Ben's holds $24 - 2t$ dollars.

Find the number of days after which the two cards hold the same amount, and what that amount is, and check the amount in both expressions. Then explain why, while both cards still hold money, the balances are equal on only one day.
1. Hint 1
  Equal balances give an equation with the variable on both sides. Collect it on the side where its coefficient is larger, even when both coefficients are negative.
2. Hint 2
  Compare $-5$ and $-2$ : the larger is $-2$ . Adding $5t$ to both sides removes the $t$ term from the left and leaves a positive coefficient on the right.
3. Hint 3
  For the last part, look at the equation you reach just before the final division, and ask how many numbers can make it true.
Answer

After $7$ days both cards hold $10$ dollars; check: $45 - 5(7) = 10$ and $24 - 2(7) = 10$ .

Full solution

The balances are equal when
$45 - 5t = 24 - 2t.$
Both variable terms are negative.
The larger coefficient is $-2$ , so keep the variable on the right and remove the $-5t$ from the left by adding $5t$ to both sides:
$45 = 24 + 3t.$

Subtract $24$ from both sides:
$21 = 3t.$
Divide both sides by $3$ :
$t = 7.$

Check in both expressions: after $7$ days Ana's card holds $45 - 5(7) = 10$ dollars and Ben's holds $24 - 2(7) = 10$ dollars.
Neither card has run out by then, so both cards hold $10$ dollars after $7$ days.

Every move above applied a property of equality to both sides, so each line has exactly the same solutions as the original equation.
The line $21 = 3t$ is true for one number only, because dividing both sides by the nonzero number $3$ leaves $t = 7$ and nothing else.
So, while both cards still hold money, the balances are equal on day $7$ and on no other day.

Answer

After $7$ days both cards hold $10$ dollars; check: $45 - 5(7) = 10$ and $24 - 2(7) = 10$ .

Key idea

With the variable on both sides, collect it where its coefficient is larger, even when both are negative, so that the coefficient left behind is positive.
Problem 7: Mark complete
Problem 8 Maya's check

Maya is solving $3(x - 2) = x + 10$ . Her work reads
$\begin{aligned} 3x - 6 &= x + 10 \\ 3x + x - 6 &= 10 \\ 4x &= 16 \\ x &= 4. \end{aligned}$
She checks $x = 4$ in her second line, where the left side comes to $10$ and matches the right side, and she stops there.

Find the first line of her work that does not follow from the line above it, and write what it should have been. Test $x = 4$ in the original equation, solve the equation correctly, and explain what Maya's successful check did and did not settle.
1. Hint 1
  Maya's check is computed correctly. Ask which equation it was carried out in, and whether that is the equation she was given.
2. Hint 2
  Removing $x$ from the right side means subtracting $x$ from both sides. Compare that with what Maya did to the left side in her second line.
3. Hint 3
  A check reports only on the equation it is carried out in. Substitute into $3(x - 2) = x + 10$ itself, working out the parentheses first.
Answer

First wrong line: her second, which should be $2x - 6 = 10$ . At $x = 4$ the original's sides are $6$ and $14$ . The solution is $x = 8$ (both sides $18$ ). Her check settled only that $4$ satisfies her second line, not the original.

Full solution

Her first line is correct, because $3(x - 2)$ equals $3x - 6$ at every number.
Her second line is where the work goes wrong.
Removing $x$ from the right side means subtracting $x$ from both sides, which lowers the left side's $3x$ to $2x$ :
$3x - 6 - x = x + 10 - x.$
Combining like terms leaves
$2x - 6 = 10.$

Maya added $x$ on the left while taking it away on the right, as if the term could jump across the equals sign and keep its sign.
The two sides changed by different amounts, so her second line is a different equation, not an equivalent one.

Test $x = 4$ in the original equation, each side on its own: the left side is $3(4 - 2) = 6$ and the right side is
$4 + 10 = 14.$
The sides differ, so $x = 4$ is not a solution.

Solve correctly from
$2x - 6 = 10.$
Add $6$ to both sides:
$2x = 16.$
Divide both sides by $2$ :
$x = 8.$
Check in the original: the left side is $3(8 - 2) = 18$ and the right side is
$8 + 10 = 18.$

Her check was honest arithmetic, and it did establish something true: $x = 4$ satisfies her second line.
But a check reports only on the equation it is carried out in.
Her second line is not equivalent to the original, so a value that passes there need not pass in the original, and here it does not.
So check in the original equation: a check in a later line can miss a slip made before that line.

Answer

First wrong line: her second, which should be $2x - 6 = 10$ . At $x = 4$ the original's sides are $6$ and $14$ . The solution is $x = 8$ (both sides $18$ ). Her check settled only that $4$ satisfies her second line, not the original.

Key idea

Check in the original equation: a check in a later line can miss a slip made before that line.
Problem 8: Mark complete
Problem 9 Four first steps

Four students each take one first step on $6(x - 4) = -2x + 16$ and write down the equation it produces.

Line A: $3(x - 4) = -x + 8$

Line B: $6x - 4 = -2x + 16$

Line C: $6(x - 4) + 2x = 16$

Line D: $3(x - 4) = -x + 16$

For each line, decide whether it has exactly the same solutions as the given equation. Where it does, name the move that produced it; where it does not, name the mistake. Then solve the given equation and check your value in it.
1. Hint 1
  A line keeps exactly the same solutions when it comes from one property of equality applied to both whole sides, or from rewriting a side as something equal to it.
2. Hint 2
  Two of the lines try to halve the equation. A product is halved by halving one factor, but every term of a sum must be halved.
3. Hint 3
  Once you have solved the given equation, substitute that value into each line: a line it fails cannot have the same solutions.
Answer

A: same (both sides divided by $2$ ). B: not (the $6$ must also multiply the $4$ , giving $6x - 24$ ). C: same ( $2x$ added to both sides). D: not (the $16$ was left unhalved). The solution is $x = 5$ , where both sides equal $6$ .

Full solution

A line has exactly the same solutions as the given equation when it comes from applying a property of equality to both whole sides, or from replacing a side with an expression equal to it at every number.

Line A divides both sides by $2$ , the division property of equality, allowed because $2$ is not zero.
The left side is a product, so halving one factor halves it: $6(x - 4)$ becomes $3(x - 4)$ .
The right side is a sum, so each of its terms is halved: $-2x + 16$ becomes $-x + 8$ .
So line A has the same solutions.

Line B distributes the $6$ to the $x$ but not to the $4$ .
The correct expansion is $6x - 24$ , and $6x - 4$ is $20$ more than that at every number.
So line B is a different equation.

Line C adds $2x$ to both sides, the addition property of equality.
On the right the $-2x$ and the added $2x$ cancel, leaving $16$ .
So line C has the same solutions; collecting before distributing is unusual, but it is legal.

Line D halves the left side correctly, but on the right it halves only the $-2x$ and leaves the $16$ whole instead of making it $8$ .
The right side was not halved as a whole, so line D is a different equation.

Now solve the given equation.
Distribute the $6$ :
$6x - 24 = -2x + 16.$
Add $2x$ to both sides:
$8x - 24 = 16.$
Add $24$ to both sides:
$8x = 40.$
Divide both sides by $8$ :
$x = 5.$

Check in the given equation: the left side is $6(5 - 4) = 6$ and the right side is
$-2(5) + 16 = 6.$
Substituting $x = 5$ into the lines agrees with the verdicts: lines A and C balance, at $3$ and $16$ , while line B gives $26$ against $6$ and line D gives $3$ against $11$ .

Answer

A: same (both sides divided by $2$ ). B: not (the $6$ must also multiply the $4$ , giving $6x - 24$ ). C: same ( $2x$ added to both sides). D: not (the $16$ was left unhalved). The solution is $x = 5$ , where both sides equal $6$ .

Key idea

A step keeps the solutions when it applies one legal move to both whole sides or rewrites a side as an equal expression; a move that reaches only part of a side can change them.
Problem 9: Mark complete
Problem 10 Putting an equation together

Start from the statement $x = -2$ . Apply exactly two properties of equality to build an equation of the form $ax + b = c$ : first multiply or divide both sides by a nonzero number, then add or subtract a number. Write the equation after each move, and name the property you used.

Explain why the equation you built has no solution other than $-2$ . Then explain why multiplying both sides by $0$ is not allowed as one of the moves.
1. Hint 1
  Building runs the solving moves forwards. Every legal move (adding or subtracting a number, multiplying or dividing by a nonzero number) can be run backwards by the opposite move.
2. Hint 2
  Multiply both sides of $x = -2$ by a nonzero number of your choice, then add a number of your choice to both sides.
3. Hint 3
  Work out what multiplying both sides of any equation by $0$ produces, and which numbers make the result true.
Answer

One build: multiply by $3$ (multiplication property), $3x = -6$ ; add $5$ (addition property), $3x + 5 = -1$ . Any nonzero multiplier and any added number work. Only $-2$ solves it. Multiplying by $0$ gives $0 = 0$ , true for every number.

Full solution

Start from the statement whose solution is to be kept, and apply each move to both sides.
Multiply both sides by $3$ , the multiplication property of equality used with the nonzero number $3$ :
$3x = -6.$
Add $5$ to both sides, the addition property of equality used with $5$ :
$3x + 5 = -1.$

The result has the form $ax + b = c$ , with $a = 3$ , $b = 5$ and $c = -1$ .
Substituting $-2$ confirms the build:
$3(-2) + 5 = -1.$
Any nonzero multiplier and any added number would do as well, so there are many correct builds.

Each move can be run backwards.
Subtracting $5$ from both sides undoes the addition, and dividing both sides by $3$ undoes the multiplication, which is allowed because $3$ is not zero.
Read forwards, the moves show that $-2$ satisfies the built equation.
Read backwards, they show that any number satisfying the built equation also satisfies $x = -2$ .
That statement is true for one number only, so the built equation is too.

Multiplying both sides of any equation by $0$ produces $0 = 0$ , which every number satisfies.
Applied to the built equation, it would gain every other number as a solution.
The step cannot be undone either, because every equation leads to the same line $0 = 0$ , so nothing tells you which equation it came from.
That is why the multiplication and division properties require a nonzero number.

Answer

One build: multiply by $3$ (multiplication property), $3x = -6$ ; add $5$ (addition property), $3x + 5 = -1$ . Any nonzero multiplier and any added number work. Only $-2$ solves it. Multiplying by $0$ gives $0 = 0$ , true for every number.

Key idea

A legal move keeps the solutions because it can be undone; multiplying by zero cannot be undone, so it is not a legal move.
Problem 10: Mark complete

← Back to lesson