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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)

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Problem 1 of 10

Problem 1 Testing a ticket

A number ticket shows $-2$ . Does it satisfy $\frac{x-1}{3}+2=1$ ?
1. Hint 1
  An equation accepts a number when both sides have the same value.
2. Hint 2
  Replace the entire $x-1$ by $-2-1$ before dividing.
Answer

Yes.

Full solution

Substitute the ticket value into the left side.
$\begin{aligned}\frac{-2-1}{3}+2&=-1+2\\&=1\end{aligned}$

The right side is also $1$ , so the ticket satisfies the equation.

Answer

Yes.

Key idea

Test a proposed solution by evaluating both sides of the original equation.
Problem 1: Mark complete
Problem 2 Inside the brackets

Solve $\frac{3(x+2)-x}{4}=5-x$ for real $x$ .
1. Hint 1
  The denominator is a nonzero constant, so it can be cleared safely.
2. Hint 2
  Multiply both sides by $4$ , then collect the terms containing $x$ .
Answer

$x=\frac73$ .

Full solution

Multiply by $4$ and distribute.
$\begin{aligned}3(x+2)-x&=20-4x\\2x+6&=20-4x\\6x&=14\\x&=\frac73\end{aligned}$

At $x=\frac73$ , each side of the original is $\frac83$ .

Answer

$x=\frac73$ .

Key idea

Clearing constant denominators can expose the linear equation underneath.
Problem 2: Mark complete
Problem 3 The folded expression

Classify $2[3x-(x-4)]=4(x+2)$ as conditional, an identity, or a contradiction.
1. Hint 1
  Compare the simplified expressions before trying to isolate a number.
2. Hint 2
  First simplify $3x-(x-4)$ , including the sign before the parentheses.
Answer

Identity; solution set $\mathbb R$ .

Full solution

Simplifying the left side gives
$\begin{aligned}2[3x-(x-4)]&=2(2x+4)\\&=4x+8\end{aligned}$

The right side also expands to $4x+8$ .
Subtracting it from both sides leaves $0=0$ , so every real number works.

Answer

Identity; solution set $\mathbb R$ .

Key idea

An identity can be hidden by different arrangements of the same expression.
Problem 3: Mark complete
Problem 4 Two packing records

Two records describe the same number of cards. One says three packets of $x$ cards plus four loose cards. The other says two packets of $x+5$ cards minus three damaged cards. Find $x$ and check that it is a possible whole-number packet size.
1. Hint 1
  The two records count the same total.
2. Hint 2
  Translate the first record as $3x+4$ and expand the second total.
Answer

$x=3$ cards per packet; both records give $13$ cards.

Full solution

Equate the recorded totals.
$\begin{aligned}3x+4&=2(x+5)-3\\3x+4&=2x+7\\x&=3\end{aligned}$

Three is a positive whole number.
The first total is $3(3)+4=13$ cards and the second is $2(3+5)-3=13$ cards.

Answer

$x=3$ cards per packet; both records give $13$ cards.

Key idea

Two descriptions of one total produce an equation whose solution must also fit the context.
Problem 4: Mark complete
Problem 5 A ratio balance

Solve $\frac{x+6}{x-2}=\frac{x-1}{x-2}+2$ over the real numbers, listing excluded values first.
1. Hint 1
  The original denominator decides which inputs are eligible.
2. Hint 2
  After excluding its zero, multiply by $x-2$ and distribute the separate $2$ .
Answer

Excluded: $x=2$ ; solution: $x=\frac{11}{2}$ .

Full solution

The denominator requires $x\ne2$ .
On this domain clearing it is reversible.
$\begin{aligned}x+6&=x-1+2(x-2)\\x+6&=3x-5\\11&=2x\\x&=\frac{11}{2}\end{aligned}$

The candidate is allowed.
The original left side is $\frac{23}{7}$ ; the right side is
$\frac97+2=\frac{23}{7}.$

Answer

Excluded: $x=2$ ; solution: $x=\frac{11}{2}$ .

Key idea

Keep the excluded values while clearing denominators, even when the candidate turns out to be allowed.
Problem 5: Mark complete
Problem 6 A missing entry

A balance record reads $5(x-2)+7=2(x+1)+3x+c$ , where $c$ is a fixed real number. Find the value of $c$ that makes the record true for every real $x$ .
1. Hint 1
  An identity must have matching constants as well as matching variable terms.
2. Hint 2
  Expand both sides and compare what remains after subtracting $5x$ .
Answer

$c=-5$ .

Full solution

The two sides simplify as follows.
$\begin{aligned}5x-3&=5x+2+c\\-3&=2+c\\c&=-5\end{aligned}$

With this value both sides equal $5x-3$ for every real $x$ .

Answer

$c=-5$ .

Key idea

When variable coefficients already match, matching the constants creates an identity.
Problem 6: Mark complete
Problem 7 A printed receipt

A receipt claims $\frac{x+8}{2}=\frac{x}{2}+5$ for a real number $x$ . Reduce this to $ax=b$ , then state its solution set and classification.
1. Hint 1
  Clear the constant denominator to compare the variable terms.
2. Hint 2
  After multiplying by $2$ , subtract $x$ from both sides.
Answer

$0x=2$ ; solution set $\varnothing$ ; contradiction.

Full solution

Multiplying by $2$ is reversible.
$\begin{aligned}x+8&=x+10\\0x&=2\end{aligned}$

The last equation is false for every real $x$ , so the original has no solution and is a contradiction.

Answer

$0x=2$ ; solution set $\varnothing$ ; contradiction.

Key idea

A nonzero constant left after every variable cancels signals a contradiction.
Problem 7: Mark complete
Problem 8 The same addition

A solver adds $x^2+1$ to both sides of a linear equation and says the new equation has exactly the original solutions. Is that claim correct? Explain.
1. Hint 1
  Judge the operation by its effect on the solution set, not by how complicated it looks.
2. Hint 2
  Ask whether the added expression exists for every real input and can be subtracted back.
Answer

Yes, the equations are equivalent.

Full solution

The expression $x^2+1$ is defined for every real $x$ .
If the original sides are $L$ and $R$ , the new equation is
$L+(x^2+1)=R+(x^2+1).$

Subtracting $x^2+1$ from both sides restores $L=R$ .
Both steps preserve equality, so neither gains nor loses a solution.

Answer

Yes, the equations are equivalent.

Key idea

Adding an expression defined throughout the domain preserves the solution set even if the written equation looks more complicated.
Problem 8: Mark complete
Problem 9 Two identical stamps

A solver multiplies both sides of $x+3=8$ by $x-5$ and says that every multiplication by a variable expression must introduce an extra solution. Decide whether that happened here and explain.
1. Hint 1
  The zero of the multiplier is the only place a new solution could appear.
2. Hint 2
  Compare that zero with the solution of the original equation.
Answer

No extra solution; both equations have solution set $\{5\}$ .

Full solution

The original gives $x=5$ .
The multiplied equation reduces to
$\begin{aligned}(x+3)(x-5)&=8(x-5)\\(x-5)^2&=0\end{aligned}$

This also holds exactly at $x=5$ .
The multiplier vanishes at a value that was already a solution, so this particular step adds nothing.
It has no general guarantee of safety.

Answer

No extra solution; both equations have solution set $\{5\}$ .

Key idea

A step that can add solutions does not necessarily add one in every equation.
Problem 9: Mark complete
Problem 10 A fractional report

The equation $\frac{x+1}{x+1}=\frac{2x+2}{2x+2}$ is reported as true for every real $x$ . Is the report accurate? Give the exact solution set and explain.
1. Hint 1
  An input must belong to the original domain before it can satisfy the equation.
2. Hint 2
  Find where either denominator is zero, then compare the fractions elsewhere.
Answer

No; solution set $\mathbb R\setminus\{-1\}$ .

Full solution

Both denominators vanish at $x=-1$ , so exclude that value.
For every other real input, each numerator equals its denominator.
$\frac{x+1}{x+1}=1$

The other fraction is also $1$ , so every allowed input solves the equation.
It is an identity on its domain, which omits $-1$ .

Answer

No; solution set $\mathbb R\setminus\{-1\}$ .

Key idea

A canceled fraction retains the domain restriction of its original denominator.
Problem 10: Mark complete

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