Standalone lesson · 50 equations

Linear Equations: 50 Questions, Step by Step

Solve for \(x\) by keeping both sides balanced. Start with a single inverse operation, then build toward parentheses, nested groups, and equations with fractions.

The equations below appear in your original order and gradually increase in difficulty. Work on paper before opening the blue Answer control. Every answer includes the algebraic steps, the reason for each operation, and substitution into the original equation.

Published October 24, 2026 · School of Math · Independent practice

School of Math cover: Linear Equations, 50 Questions, Step by Step, with equations x + 5 = 12, 3x − 4 = 17, and 5x + 2 = 2x + 20 rising on green steps

This lesson stands on its own. It is not assigned to any course, class number, or syllabus, and it does not require attendance at an earlier lesson.

Questions 1–10

One-step equations

Undo one operation at a time

An equation states that two expressions are equal. Solving means finding the value of \(x\) that makes the statement true. These equations are linear: after simplifying, the variable has power 1, with no products of variable terms.

Keep the equation balanced by doing the same operation to both sides. Addition and subtraction undo each other; multiplication and division by the same nonzero number undo each other. For \(x+5=12\), subtract 5 from both sides.

The sign belongs to the term: \(-x\) means \((-1)x\). If you reach \(-x=-11\), divide both sides by \(-1\).

Question 1

\[x + 5 = 12\]
Answer

Solution: \(x = 7\)

  1. Subtract 5 from both sides to undo adding 5.

    \[x = 12 - 5\]

  2. Simplify the right side.

    \[x = 7\]

Check in the original equation

Substitute \(x = 7\) and evaluate each side separately.

Left side

\[\left(7\right) + 5 = 12\]

Right side

\[12 = 12\]

Both sides equal \(12\), so \(x = 7\) is correct.

Question 2

\[x - 8 = 11\]
Answer

Solution: \(x = 19\)

  1. Add 8 to both sides to undo subtracting 8.

    \[x = 11 + 8\]

  2. Simplify.

    \[x = 19\]

Check in the original equation

Substitute \(x = 19\) and evaluate each side separately.

Left side

\[\left(19\right) - 8 = 11\]

Right side

\[11 = 11\]

Both sides equal \(11\), so \(x = 19\) is correct.

Question 3

\[x + 14 = 30\]
Answer

Solution: \(x = 16\)

  1. Subtract 14 from both sides.

    \[x = 30 - 14\]

  2. Simplify.

    \[x = 16\]

Check in the original equation

Substitute \(x = 16\) and evaluate each side separately.

Left side

\[\left(16\right) + 14 = 30\]

Right side

\[30 = 30\]

Both sides equal \(30\), so \(x = 16\) is correct.

Question 4

\[3x = 21\]
Answer

Solution: \(x = 7\)

  1. Divide both sides by 3, the coefficient of x.

    \[x = \frac{21}{3}\]

  2. Simplify.

    \[x = 7\]

Check in the original equation

Substitute \(x = 7\) and evaluate each side separately.

Left side

\[3\left(7\right) = 21\]

Right side

\[21 = 21\]

Both sides equal \(21\), so \(x = 7\) is correct.

Question 5

\[\dfrac{x}{4} = 6\]
Answer

Solution: \(x = 24\)

  1. Multiply both sides by 4 to undo division by 4.

    \[x = 4 \cdot 6\]

  2. Simplify.

    \[x = 24\]

Check in the original equation

Substitute \(x = 24\) and evaluate each side separately.

Left side

\[\dfrac{\left(24\right)}{4} = 6\]

Right side

\[6 = 6\]

Both sides equal \(6\), so \(x = 24\) is correct.

Question 6

\[7x = 56\]
Answer

Solution: \(x = 8\)

  1. Divide both sides by 7.

    \[x = \frac{56}{7}\]

  2. Simplify.

    \[x = 8\]

Check in the original equation

Substitute \(x = 8\) and evaluate each side separately.

Left side

\[7\left(8\right) = 56\]

Right side

\[56 = 56\]

Both sides equal \(56\), so \(x = 8\) is correct.

Question 7

\[x - 12 = -5\]
Answer

Solution: \(x = 7\)

  1. Add 12 to both sides. A negative right side is handled by the same equality rule.

    \[x = -5 + 12\]

  2. Simplify.

    \[x = 7\]

Check in the original equation

Substitute \(x = 7\) and evaluate each side separately.

Left side

\[\left(7\right) - 12 = -5\]

Right side

\[-5 = -5\]

Both sides equal \(-5\), so \(x = 7\) is correct.

Question 8

\[5x = -35\]
Answer

Solution: \(x = -7\)

  1. Divide both sides by 5. A negative number divided by a positive number is negative.

    \[x = - \frac{35}{5}\]

  2. Simplify.

    \[x = -7\]

Check in the original equation

Substitute \(x = -7\) and evaluate each side separately.

Left side

\[5\left(-7\right) = -35\]

Right side

\[-35 = -35\]

Both sides equal \(-35\), so \(x = -7\) is correct.

Question 9

\[\dfrac{x}{6} = -3\]
Answer

Solution: \(x = -18\)

  1. Multiply both sides by 6.

    \[x = 6 \left(-3\right)\]

  2. Simplify.

    \[x = -18\]

Check in the original equation

Substitute \(x = -18\) and evaluate each side separately.

Left side

\[\dfrac{\left(-18\right)}{6} = -3\]

Right side

\[-3 = -3\]

Both sides equal \(-3\), so \(x = -18\) is correct.

Question 10

\[18 - x = 7\]
Answer

Solution: \(x = 11\)

  1. Subtract 18 from both sides. The variable term is negative x, not x.

    \[- x = 7 - 18\]

  2. Simplify the right side.

    \[- x = -11\]

  3. Divide both sides by negative 1.

    \[x = 11\]

Check in the original equation

Substitute \(x = 11\) and evaluate each side separately.

Left side

\[18 - \left(11\right) = 7\]

Right side

\[7 = 7\]

Both sides equal \(7\), so \(x = 11\) is correct.

Questions 11–18

Two-step equations

Undo the outside operation first

For an equation such as \(2x+5=19\), subtract the constant term first, then divide by the coefficient of \(x\). This reverses the order of the operations that formed the expression.

A fraction like \(\frac{x}{3}\) means “\(x\) divided by 3.” Once that fraction is isolated, multiply both sides by 3. Keep negative signs attached to terms throughout the calculation.

Question 11

\[2x + 5 = 19\]
Answer

Solution: \(x = 7\)

  1. Subtract 5 from both sides before undoing the multiplication.

    \[2 x = 19 - 5\]

  2. Simplify.

    \[2 x = 14\]

  3. Divide both sides by 2.

    \[x = 7\]

Check in the original equation

Substitute \(x = 7\) and evaluate each side separately.

Left side

\[2\left(7\right) + 5 = 19\]

Right side

\[19 = 19\]

Both sides equal \(19\), so \(x = 7\) is correct.

Question 12

\[3x - 7 = 20\]
Answer

Solution: \(x = 9\)

  1. Add 7 to both sides.

    \[3 x = 20 + 7\]

  2. Simplify.

    \[3 x = 27\]

  3. Divide both sides by 3.

    \[x = 9\]

Check in the original equation

Substitute \(x = 9\) and evaluate each side separately.

Left side

\[3\left(9\right) - 7 = 20\]

Right side

\[20 = 20\]

Both sides equal \(20\), so \(x = 9\) is correct.

Question 13

\[5x + 8 = 43\]
Answer

Solution: \(x = 7\)

  1. Subtract 8 from both sides.

    \[5 x = 43 - 8\]

  2. Simplify.

    \[5 x = 35\]

  3. Divide both sides by 5.

    \[x = 7\]

Check in the original equation

Substitute \(x = 7\) and evaluate each side separately.

Left side

\[5\left(7\right) + 8 = 43\]

Right side

\[43 = 43\]

Both sides equal \(43\), so \(x = 7\) is correct.

Question 14

\[4x - 9 = -25\]
Answer

Solution: \(x = -4\)

  1. Add 9 to both sides.

    \[4 x = -25 + 9\]

  2. Simplify.

    \[4 x = -16\]

  3. Divide both sides by 4.

    \[x = -4\]

Check in the original equation

Substitute \(x = -4\) and evaluate each side separately.

Left side

\[4\left(-4\right) - 9 = -25\]

Right side

\[-25 = -25\]

Both sides equal \(-25\), so \(x = -4\) is correct.

Question 15

\[12 - 3x = 27\]
Answer

Solution: \(x = -5\)

  1. Subtract 12 from both sides. Keep the negative sign with the coefficient 3.

    \[- 3 x = 27 - 12\]

  2. Simplify.

    \[- 3 x = 15\]

  3. Divide both sides by negative 3.

    \[x = -5\]

Check in the original equation

Substitute \(x = -5\) and evaluate each side separately.

Left side

\[12 - 3\left(-5\right) = 27\]

Right side

\[27 = 27\]

Both sides equal \(27\), so \(x = -5\) is correct.

Question 16

\[\dfrac{x}{3} + 4 = 10\]
Answer

Solution: \(x = 18\)

  1. Subtract 4 from both sides.

    \[\frac{x}{3} = 10 - 4\]

  2. Simplify.

    \[\frac{x}{3} = 6\]

  3. Multiply both sides by 3.

    \[x = 18\]

Check in the original equation

Substitute \(x = 18\) and evaluate each side separately.

Left side

\[\dfrac{\left(18\right)}{3} + 4 = 10\]

Right side

\[10 = 10\]

Both sides equal \(10\), so \(x = 18\) is correct.

Question 17

\[\dfrac{x}{5} - 6 = 2\]
Answer

Solution: \(x = 40\)

  1. Add 6 to both sides.

    \[\frac{x}{5} = 2 + 6\]

  2. Simplify.

    \[\frac{x}{5} = 8\]

  3. Multiply both sides by 5.

    \[x = 40\]

Check in the original equation

Substitute \(x = 40\) and evaluate each side separately.

Left side

\[\dfrac{\left(40\right)}{5} - 6 = 2\]

Right side

\[2 = 2\]

Both sides equal \(2\), so \(x = 40\) is correct.

Question 18

\[7 - \dfrac{x}{2} = 12\]
Answer

Solution: \(x = -10\)

  1. Subtract 7 from both sides. The entire fraction is negative.

    \[\frac{\left(-1\right) x}{2} = 12 - 7\]

  2. Simplify.

    \[\frac{\left(-1\right) x}{2} = 5\]

  3. Multiply both sides by negative 2.

    \[x = -10\]

Check in the original equation

Substitute \(x = -10\) and evaluate each side separately.

Left side

\[7 - \dfrac{\left(-10\right)}{2} = 12\]

Right side

\[12 = 12\]

Both sides equal \(12\), so \(x = -10\) is correct.

Questions 19–20

Variables on both sides

Collect terms without changing the balance

Subtract a variable term from both sides to place all the \(x\)-terms on one side. Then use inverse operations to isolate \(x\). “Moving a term” is shorthand for adding or subtracting that term on both sides.

Combine like terms only: \(6x-2x=4x\), but \(4x+11\) cannot be combined into \(15x\). If variable terms ever cancel completely, a true statement means infinitely many solutions and a false statement means no solution. Each of these 50 questions has exactly one solution.

Question 19

\[6x + 11 = 2x + 35\]
Answer

Solution: \(x = 6\)

  1. Subtract 2x from both sides to collect the variable terms on the left.

    \[4 x + 11 = 35\]

  2. Subtract 11 from both sides.

    \[4 x = 24\]

  3. Divide both sides by 4.

    \[x = 6\]

Check in the original equation

Substitute \(x = 6\) and evaluate each side separately.

Left side

\[6\left(6\right) + 11 = 47\]

Right side

\[2\left(6\right) + 35 = 47\]

Both sides equal \(47\), so \(x = 6\) is correct.

Question 20

\[9x - 13 = 5x + 15\]
Answer

Solution: \(x = 7\)

  1. Subtract 5x from both sides.

    \[4 x - 13 = 15\]

  2. Add 13 to both sides.

    \[4 x = 28\]

  3. Divide both sides by 4.

    \[x = 7\]

Check in the original equation

Substitute \(x = 7\) and evaluate each side separately.

Left side

\[9\left(7\right) - 13 = 50\]

Right side

\[5\left(7\right) + 15 = 50\]

Both sides equal \(50\), so \(x = 7\) is correct.

Questions 21–30

Parentheses and distribution

Multiply every term in the group

The distributive property says \(a(b+c)=ab+ac\). Apply the factor to every term, including its sign. For example, \(-3(x-4)=-3x+12\), not \(-3x-12\).

You can sometimes avoid distributing: in \(3(x+4)=27\), divide both sides by 3 first. Choose a valid operation that makes the equation simpler, then combine like terms and isolate \(x\).

Question 21

\[3(x + 4) = 27\]
Answer

Solution: \(x = 5\)

  1. Divide both sides by 3 first; this removes the outside factor from the entire group.

    \[x + 4 = 9\]

  2. Subtract 4 from both sides.

    \[x = 5\]

Check in the original equation

Substitute \(x = 5\) and evaluate each side separately.

Left side

\[3(\left(5\right) + 4) = 27\]

Right side

\[27 = 27\]

Both sides equal \(27\), so \(x = 5\) is correct.

Question 22

\[5(x - 2) = 35\]
Answer

Solution: \(x = 9\)

  1. Divide both sides by 5.

    \[x - 2 = 7\]

  2. Add 2 to both sides.

    \[x = 9\]

Check in the original equation

Substitute \(x = 9\) and evaluate each side separately.

Left side

\[5(\left(9\right) - 2) = 35\]

Right side

\[35 = 35\]

Both sides equal \(35\), so \(x = 9\) is correct.

Question 23

\[2(x + 6) - 5 = 21\]
Answer

Solution: \(x = 7\)

  1. Distribute the 2 to both terms inside the parentheses.

    \[2 x + 12 - 5 = 21\]

  2. Combine the constant terms: 12 minus 5 is 7.

    \[2 x + 7 = 21\]

  3. Subtract 7 from both sides.

    \[2 x = 14\]

  4. Divide both sides by 2.

    \[x = 7\]

Check in the original equation

Substitute \(x = 7\) and evaluate each side separately.

Left side

\[2(\left(7\right) + 6) - 5 = 21\]

Right side

\[21 = 21\]

Both sides equal \(21\), so \(x = 7\) is correct.

Question 24

\[4(x - 3) + 7 = 31\]
Answer

Solution: \(x = 9\)

  1. Distribute 4 to x and to negative 3.

    \[4 x - 12 + 7 = 31\]

  2. Combine the constants.

    \[4 x - 5 = 31\]

  3. Add 5 to both sides.

    \[4 x = 36\]

  4. Divide both sides by 4.

    \[x = 9\]

Check in the original equation

Substitute \(x = 9\) and evaluate each side separately.

Left side

\[4(\left(9\right) - 3) + 7 = 31\]

Right side

\[31 = 31\]

Both sides equal \(31\), so \(x = 9\) is correct.

Question 25

\[6 - 2(x + 1) = 14\]
Answer

Solution: \(x = -5\)

  1. Distribute negative 2 to both terms. The constant contribution is negative 2, not positive 2.

    \[6 - 2 x - 2 = 14\]

  2. Combine the constants.

    \[4 - 2 x = 14\]

  3. Subtract 4 from both sides.

    \[- 2 x = 10\]

  4. Divide both sides by negative 2.

    \[x = -5\]

Check in the original equation

Substitute \(x = -5\) and evaluate each side separately.

Left side

\[6 - 2(\left(-5\right) + 1) = 14\]

Right side

\[14 = 14\]

Both sides equal \(14\), so \(x = -5\) is correct.

Question 26

\[3(x + 5) = 2x + 24\]
Answer

Solution: \(x = 9\)

  1. Distribute 3 on the left.

    \[3 x + 15 = 2 x + 24\]

  2. Subtract 2x from both sides.

    \[x + 15 = 24\]

  3. Subtract 15 from both sides.

    \[x = 9\]

Check in the original equation

Substitute \(x = 9\) and evaluate each side separately.

Left side

\[3(\left(9\right) + 5) = 42\]

Right side

\[2\left(9\right) + 24 = 42\]

Both sides equal \(42\), so \(x = 9\) is correct.

Question 27

\[4(x - 2) = 2(x + 7)\]
Answer

Solution: \(x = 11\)

  1. Distribute the outside factors on both sides.

    \[4 x - 8 = 2 x + 14\]

  2. Subtract 2x from both sides.

    \[2 x - 8 = 14\]

  3. Add 8 to both sides.

    \[2 x = 22\]

  4. Divide both sides by 2.

    \[x = 11\]

Check in the original equation

Substitute \(x = 11\) and evaluate each side separately.

Left side

\[4(\left(11\right) - 2) = 36\]

Right side

\[2(\left(11\right) + 7) = 36\]

Both sides equal \(36\), so \(x = 11\) is correct.

Question 28

\[5(x + 3) - 2x = 36\]
Answer

Solution: \(x = 7\)

  1. Distribute 5.

    \[5 x + 15 - 2 x = 36\]

  2. Combine the x terms on the same side.

    \[3 x + 15 = 36\]

  3. Subtract 15 from both sides.

    \[3 x = 21\]

  4. Divide both sides by 3.

    \[x = 7\]

Check in the original equation

Substitute \(x = 7\) and evaluate each side separately.

Left side

\[5(\left(7\right) + 3) - 2\left(7\right) = 36\]

Right side

\[36 = 36\]

Both sides equal \(36\), so \(x = 7\) is correct.

Question 29

\[7x - 3(x - 4) = 40\]
Answer

Solution: \(x = 7\)

  1. Distribute negative 3. Negative 3 times negative 4 gives positive 12.

    \[7 x - 3 x + 12 = 40\]

  2. Combine the x terms.

    \[4 x + 12 = 40\]

  3. Subtract 12 from both sides.

    \[4 x = 28\]

  4. Divide both sides by 4.

    \[x = 7\]

Check in the original equation

Substitute \(x = 7\) and evaluate each side separately.

Left side

\[7\left(7\right) - 3(\left(7\right) - 4) = 40\]

Right side

\[40 = 40\]

Both sides equal \(40\), so \(x = 7\) is correct.

Question 30

\[2(3x - 5) + 4 = 4x + 18\]
Answer

Solution: \(x = 12\)

  1. Distribute 2 to both terms in the parentheses.

    \[6 x - 10 + 4 = 4 x + 18\]

  2. Combine the constants on the left.

    \[6 x - 6 = 4 x + 18\]

  3. Subtract 4x from both sides.

    \[2 x - 6 = 18\]

  4. Add 6 to both sides.

    \[2 x = 24\]

  5. Divide both sides by 2.

    \[x = 12\]

Check in the original equation

Substitute \(x = 12\) and evaluate each side separately.

Left side

\[2(3\left(12\right) - 5) + 4 = 66\]

Right side

\[4\left(12\right) + 18 = 66\]

Both sides equal \(66\), so \(x = 12\) is correct.

Questions 31–37

Multiple groups and nested brackets

Simplify carefully, from the inside out

Brackets and parentheses both group expressions. Simplify an inner group before distributing an outer factor, or remove an outside factor from both sides when convenient.

Write a separate line for distribution and for combining like terms. In \(5-2[3(x+1)-4]\), the factor outside the bracket is \(-2\); it multiplies the entire bracket.

Question 31

\[5(x - 2) - 3(x + 4) = 8\]
Answer

Solution: \(x = 15\)

  1. Distribute 5 and negative 3. In particular, negative 3 times positive 4 is negative 12.

    \[5 x - 10 - 3 x - 12 = 8\]

  2. Combine the x terms and the constants.

    \[2 x - 22 = 8\]

  3. Add 22 to both sides.

    \[2 x = 30\]

  4. Divide both sides by 2.

    \[x = 15\]

Check in the original equation

Substitute \(x = 15\) and evaluate each side separately.

Left side

\[5(\left(15\right) - 2) - 3(\left(15\right) + 4) = 8\]

Right side

\[8 = 8\]

Both sides equal \(8\), so \(x = 15\) is correct.

Question 32

\[8 - 3(2x - 5) = 2x - 17\]
Answer

Solution: \(x = 5\)

  1. Distribute negative 3: the two products are negative 6x and positive 15.

    \[8 - 6 x + 15 = 2 x - 17\]

  2. Combine the constants on the left.

    \[23 - 6 x = 2 x - 17\]

  3. Add 6x to both sides.

    \[23 = 8 x - 17\]

  4. Add 17 to both sides.

    \[40 = 8 x\]

  5. Divide both sides by 8 and write x on the left.

    \[x = 5\]

Check in the original equation

Substitute \(x = 5\) and evaluate each side separately.

Left side

\[8 - 3(2\left(5\right) - 5) = -7\]

Right side

\[2\left(5\right) - 17 = -7\]

Both sides equal \(-7\), so \(x = 5\) is correct.

Question 33

\[4(2x + 1) - 5(x - 3) = 40\]
Answer

Solution: \(x = 7\)

  1. Distribute 4 and negative 5; the last product is positive 15.

    \[8 x + 4 - 5 x + 15 = 40\]

  2. Combine like terms.

    \[3 x + 19 = 40\]

  3. Subtract 19 from both sides.

    \[3 x = 21\]

  4. Divide both sides by 3.

    \[x = 7\]

Check in the original equation

Substitute \(x = 7\) and evaluate each side separately.

Left side

\[4(2\left(7\right) + 1) - 5(\left(7\right) - 3) = 40\]

Right side

\[40 = 40\]

Both sides equal \(40\), so \(x = 7\) is correct.

Question 34

\[6(x - 2) + 3 = 4(x + 5) - 7\]
Answer

Solution: \(x = 11\)

  1. Distribute 6 on the left and 4 on the right.

    \[6 x - 12 + 3 = 4 x + 20 - 7\]

  2. Combine constants on each side separately.

    \[6 x - 9 = 4 x + 13\]

  3. Subtract 4x from both sides.

    \[2 x - 9 = 13\]

  4. Add 9 to both sides.

    \[2 x = 22\]

  5. Divide both sides by 2.

    \[x = 11\]

Check in the original equation

Substitute \(x = 11\) and evaluate each side separately.

Left side

\[6(\left(11\right) - 2) + 3 = 57\]

Right side

\[4(\left(11\right) + 5) - 7 = 57\]

Both sides equal \(57\), so \(x = 11\) is correct.

Question 35

\[3[2(x + 1) - 5] = 27\]
Answer

Solution: \(x = 6\)

  1. Divide both sides by 3 to remove the outermost factor.

    \[2 \left(x + 1\right) - 5 = 9\]

  2. Add 5 to both sides.

    \[2 \left(x + 1\right) = 14\]

  3. Divide both sides by 2.

    \[x + 1 = 7\]

  4. Subtract 1 from both sides.

    \[x = 6\]

Check in the original equation

Substitute \(x = 6\) and evaluate each side separately.

Left side

\[3[2(\left(6\right) + 1) - 5] = 27\]

Right side

\[27 = 27\]

Both sides equal \(27\), so \(x = 6\) is correct.

Question 36

\[2[3(x - 2) + 4] = 5x + 9\]
Answer

Solution: \(x = 13\)

  1. Work inside the brackets first: distribute 3.

    \[2 \left(3 x - 6 + 4\right) = 5 x + 9\]

  2. Combine the constants inside the brackets.

    \[2 \left(3 x - 2\right) = 5 x + 9\]

  3. Distribute the outside factor 2.

    \[6 x - 4 = 5 x + 9\]

  4. Subtract 5x from both sides.

    \[x - 4 = 9\]

  5. Add 4 to both sides.

    \[x = 13\]

Check in the original equation

Substitute \(x = 13\) and evaluate each side separately.

Left side

\[2[3(\left(13\right) - 2) + 4] = 74\]

Right side

\[5\left(13\right) + 9 = 74\]

Both sides equal \(74\), so \(x = 13\) is correct.

Question 37

\[5 - 2[3(x + 1) - 4] = 19\]
Answer

Solution: \(x = -2\)

  1. First distribute 3 inside the brackets.

    \[5 - 2 \left(3 x + 3 - 4\right) = 19\]

  2. Combine the constants inside the brackets.

    \[5 - 2 \left(3 x - 1\right) = 19\]

  3. Distribute negative 2 to both terms in the remaining group.

    \[5 - 6 x + 2 = 19\]

  4. Combine the constants.

    \[7 - 6 x = 19\]

  5. Subtract 7 from both sides.

    \[- 6 x = 12\]

  6. Divide both sides by negative 6.

    \[x = -2\]

Check in the original equation

Substitute \(x = -2\) and evaluate each side separately.

Left side

\[5 - 2[3(\left(-2\right) + 1) - 4] = 19\]

Right side

\[19 = 19\]

Both sides equal \(19\), so \(x = -2\) is correct.

Questions 38–44

Equations with fractions

Clear denominators in every term

A fraction bar groups the entire numerator. Multiplying \(\frac{x+5}{3}=7\) by 3 gives \(x+5=21\), not \(x+\frac{5}{3}=21\).

With more than one denominator, multiply every term on both sides by the least common denominator (LCD). For denominators 3 and 4, use 12. Preserve numerator groups until you distribute, especially after a subtraction sign.

All denominators here are nonzero constants, so clearing them does not exclude any value of \(x\). Answers need not be whole numbers; keep fractions exact instead of rounding.

Question 38

\[\dfrac{x + 5}{3} = 7\]
Answer

Solution: \(x = 16\)

  1. Multiply both sides by 3. The whole numerator, x plus 5, remains.

    \[x + 5 = 21\]

  2. Subtract 5 from both sides.

    \[x = 16\]

Check in the original equation

Substitute \(x = 16\) and evaluate each side separately.

Left side

\[\dfrac{\left(16\right) + 5}{3} = 7\]

Right side

\[7 = 7\]

Both sides equal \(7\), so \(x = 16\) is correct.

Question 39

\[\dfrac{2x - 3}{5} = 9\]
Answer

Solution: \(x = 24\)

  1. Multiply both sides by 5.

    \[2 x - 3 = 45\]

  2. Add 3 to both sides.

    \[2 x = 48\]

  3. Divide both sides by 2.

    \[x = 24\]

Check in the original equation

Substitute \(x = 24\) and evaluate each side separately.

Left side

\[\dfrac{2\left(24\right) - 3}{5} = 9\]

Right side

\[9 = 9\]

Both sides equal \(9\), so \(x = 24\) is correct.

Question 40

\[\dfrac{3x + 4}{2} - 5 = 12\]
Answer

Solution: \(x = 10\)

  1. Add 5 to both sides first.

    \[\frac{3 x + 4}{2} = 17\]

  2. Multiply both sides by 2.

    \[3 x + 4 = 34\]

  3. Subtract 4 from both sides.

    \[3 x = 30\]

  4. Divide both sides by 3.

    \[x = 10\]

Check in the original equation

Substitute \(x = 10\) and evaluate each side separately.

Left side

\[\dfrac{3\left(10\right) + 4}{2} - 5 = 12\]

Right side

\[12 = 12\]

Both sides equal \(12\), so \(x = 10\) is correct.

Question 41

\[\dfrac{x - 2}{3} + \dfrac{x + 4}{2} = 13\]
Answer

Solution: \(x = 14\)

  1. The least common denominator is 6. Multiply every term on both sides by 6; the multipliers of the numerators are 2 and 3.

    \[2 \left(x - 2\right) + 3 \left(x + 4\right) = 78\]

  2. Distribute both factors.

    \[2 x - 4 + 3 x + 12 = 78\]

  3. Combine like terms.

    \[5 x + 8 = 78\]

  4. Subtract 8 from both sides.

    \[5 x = 70\]

  5. Divide both sides by 5.

    \[x = 14\]

Check in the original equation

Substitute \(x = 14\) and evaluate each side separately.

Left side

\[\dfrac{\left(14\right) - 2}{3} + \dfrac{\left(14\right) + 4}{2} = 13\]

Right side

\[13 = 13\]

Both sides equal \(13\), so \(x = 14\) is correct.

Question 42

\[\dfrac{2x + 1}{3} - \dfrac{x - 2}{4} = 5\]
Answer

Solution: \(x = 10\)

  1. Multiply every term by the least common denominator, 12. Keep the subtraction in front of the entire second numerator.

    \[4 \left(2 x + 1\right) - 3 \left(x - 2\right) = 60\]

  2. Distribute 4 and negative 3; subtracting 3 times negative 2 gives positive 6.

    \[8 x + 4 - 3 x + 6 = 60\]

  3. Combine like terms.

    \[5 x + 10 = 60\]

  4. Subtract 10 from both sides.

    \[5 x = 50\]

  5. Divide both sides by 5.

    \[x = 10\]

Check in the original equation

Substitute \(x = 10\) and evaluate each side separately.

Left side

\[\dfrac{2\left(10\right) + 1}{3} - \dfrac{\left(10\right) - 2}{4} = 5\]

Right side

\[5 = 5\]

Both sides equal \(5\), so \(x = 10\) is correct.

Question 43

\[\dfrac{3x - 5}{2} = \dfrac{x + 7}{3}\]
Answer

Solution: \(x = \frac{29}{7}\)

  1. Multiply both sides by 6, the least common denominator.

    \[3 \left(3 x - 5\right) = 2 \left(x + 7\right)\]

  2. Distribute on both sides.

    \[9 x - 15 = 2 x + 14\]

  3. Subtract 2x from both sides.

    \[7 x - 15 = 14\]

  4. Add 15 to both sides.

    \[7 x = 29\]

  5. Divide both sides by 7. Leave the non-integer answer as an exact fraction.

    \[x = \frac{29}{7}\]

Check in the original equation

Substitute \(x = \frac{29}{7}\) and evaluate each side separately.

Left side

\[\dfrac{3\left(\frac{29}{7}\right) - 5}{2} = \frac{26}{7}\]

Right side

\[\dfrac{\left(\frac{29}{7}\right) + 7}{3} = \frac{26}{7}\]

Both sides equal \(\frac{26}{7}\), so \(x = \frac{29}{7}\) is correct.

Question 44

\[\dfrac{5x + 2}{4} - \dfrac{2x - 1}{3} = 5\]
Answer

Solution: \(x = \frac{50}{7}\)

  1. Multiply every term by 12. The first numerator receives a factor of 3; the second receives a factor of negative 4.

    \[3 \left(5 x + 2\right) - 4 \left(2 x - 1\right) = 60\]

  2. Distribute carefully.

    \[15 x + 6 - 8 x + 4 = 60\]

  3. Combine like terms.

    \[7 x + 10 = 60\]

  4. Subtract 10 from both sides.

    \[7 x = 50\]

  5. Divide both sides by 7 and retain the exact fraction.

    \[x = \frac{50}{7}\]

Check in the original equation

Substitute \(x = \frac{50}{7}\) and evaluate each side separately.

Left side

\[\dfrac{5\left(\frac{50}{7}\right) + 2}{4} - \dfrac{2\left(\frac{50}{7}\right) - 1}{3} = 5\]

Right side

\[5 = 5\]

Both sides equal \(5\), so \(x = \frac{50}{7}\) is correct.

Questions 45–50

Fraction challenges

Combine the methods, then verify

Identify every denominator, clear all fractions, simplify grouped expressions, collect variable terms, and isolate \(x\). Multiply standalone constants and every term on the other side by the LCD too.

Finally, substitute your answer into the original equation, not just your last simplified line. Calculate the left side and the right side independently. Matching values verify the solution.

Question 45

\[\dfrac{x + 3}{2} + \dfrac{2x - 1}{5} = x - 1\]
Answer

Solution: \(x = 23\)

  1. Multiply every term by 10. Multiply both terms on the right, not just x.

    \[5 \left(x + 3\right) + 2 \left(2 x - 1\right) = 10 \left(x - 1\right)\]

  2. Distribute all factors.

    \[5 x + 15 + 4 x - 2 = 10 x - 10\]

  3. Combine like terms on the left.

    \[9 x + 13 = 10 x - 10\]

  4. Subtract 9x from both sides.

    \[13 = x - 10\]

  5. Add 10 to both sides and write x on the left.

    \[x = 23\]

Check in the original equation

Substitute \(x = 23\) and evaluate each side separately.

Left side

\[\dfrac{\left(23\right) + 3}{2} + \dfrac{2\left(23\right) - 1}{5} = 22\]

Right side

\[\left(23\right) - 1 = 22\]

Both sides equal \(22\), so \(x = 23\) is correct.

Question 46

\[3 - \dfrac{2(x - 4)}{5} = \dfrac{x + 1}{2}\]
Answer

Solution: \(x = \frac{41}{9}\)

  1. Multiply every term by 10. The standalone 3 becomes 30. The second term becomes negative 4 times the group x minus 4.

    \[30 - 4 \left(x - 4\right) = 5 \left(x + 1\right)\]

  2. Distribute negative 4 and positive 5.

    \[30 - 4 x + 16 = 5 x + 5\]

  3. Combine constants on the left.

    \[46 - 4 x = 5 x + 5\]

  4. Add 4x to both sides.

    \[46 = 9 x + 5\]

  5. Subtract 5 from both sides.

    \[41 = 9 x\]

  6. Divide both sides by 9 and write the exact answer.

    \[x = \frac{41}{9}\]

Check in the original equation

Substitute \(x = \frac{41}{9}\) and evaluate each side separately.

Left side

\[3 - \dfrac{2(\left(\frac{41}{9}\right) - 4)}{5} = \frac{25}{9}\]

Right side

\[\dfrac{\left(\frac{41}{9}\right) + 1}{2} = \frac{25}{9}\]

Both sides equal \(\frac{25}{9}\), so \(x = \frac{41}{9}\) is correct.

Question 47

\[\dfrac{3(x - 2)}{4} + \dfrac{x + 5}{6} = 9\]
Answer

Solution: \(x = \frac{116}{11}\)

  1. Multiply every term by 12. The first numerator already contains a factor of 3, so its total coefficient becomes 9.

    \[9 \left(x - 2\right) + 2 \left(x + 5\right) = 108\]

  2. Distribute both factors.

    \[9 x - 18 + 2 x + 10 = 108\]

  3. Combine like terms.

    \[11 x - 8 = 108\]

  4. Add 8 to both sides.

    \[11 x = 116\]

  5. Divide both sides by 11.

    \[x = \frac{116}{11}\]

Check in the original equation

Substitute \(x = \frac{116}{11}\) and evaluate each side separately.

Left side

\[\dfrac{3(\left(\frac{116}{11}\right) - 2)}{4} + \dfrac{\left(\frac{116}{11}\right) + 5}{6} = 9\]

Right side

\[9 = 9\]

Both sides equal \(9\), so \(x = \frac{116}{11}\) is correct.

Question 48

\[\dfrac{2(x + 3) - 5}{3} - \dfrac{x - 4}{2} = 4\]
Answer

Solution: \(x = 10\)

  1. Multiply every term by 6. Preserve the entire first numerator as a group.

    \[2 \left(2 \left(x + 3\right) - 5\right) - 3 \left(x - 4\right) = 24\]

  2. Inside the first group, distribute 2 and combine 6 minus 5 to get 1.

    \[2 \left(2 x + 1\right) - 3 \left(x - 4\right) = 24\]

  3. Distribute the outside factors.

    \[4 x + 2 - 3 x + 12 = 24\]

  4. Combine like terms.

    \[x + 14 = 24\]

  5. Subtract 14 from both sides.

    \[x = 10\]

Check in the original equation

Substitute \(x = 10\) and evaluate each side separately.

Left side

\[\dfrac{2(\left(10\right) + 3) - 5}{3} - \dfrac{\left(10\right) - 4}{2} = 4\]

Right side

\[4 = 4\]

Both sides equal \(4\), so \(x = 10\) is correct.

Question 49

\[\dfrac{3x - 2}{4} - \dfrac{2(x + 1)}{3} = \dfrac{x - 10}{6}\]
Answer

Solution: \(x = 6\)

  1. Multiply every term by 12. In the second term, 12 divided by 3 times the existing factor 2 gives 8.

    \[3 \left(3 x - 2\right) - 8 \left(x + 1\right) = 2 \left(x - 10\right)\]

  2. Distribute all factors, including the negative 8.

    \[9 x - 6 - 8 x - 8 = 2 x - 20\]

  3. Combine like terms on the left.

    \[x - 14 = 2 x - 20\]

  4. Subtract x from both sides.

    \[-14 = x - 20\]

  5. Add 20 to both sides and write x on the left.

    \[x = 6\]

Check in the original equation

Substitute \(x = 6\) and evaluate each side separately.

Left side

\[\dfrac{3\left(6\right) - 2}{4} - \dfrac{2(\left(6\right) + 1)}{3} = - \frac{2}{3}\]

Right side

\[\dfrac{\left(6\right) - 10}{6} = - \frac{2}{3}\]

Both sides equal \(- \frac{2}{3}\), so \(x = 6\) is correct.

Question 50

\[\dfrac{2[3(x - 1) + 4]}{5} - \dfrac{3(x + 2) - 1}{4} = \dfrac{x + 13}{10}\]
Answer

Solution: \(x = \frac{43}{7}\)

  1. The least common denominator of 5, 4, and 10 is 20. Multiply every term by 20. The first numerator already has a factor of 2, giving a total factor of 8.

    \[\begin{aligned}&8 \left(3 \left(x - 1\right) + 4\right) - 5 \left(3 \left(x + 2\right) - 1\right) \\&= 2 \left(x + 13\right)\end{aligned}\]

  2. Simplify inside each bracket: 3(x minus 1) plus 4 becomes 3x plus 1, and 3(x plus 2) minus 1 becomes 3x plus 5.

    \[8 \left(3 x + 1\right) - 5 \left(3 x + 5\right) = 2 \left(x + 13\right)\]

  3. Distribute 8, negative 5, and 2 to every term in their groups.

    \[24 x + 8 - 15 x - 25 = 2 x + 26\]

  4. Combine like terms on the left.

    \[9 x - 17 = 2 x + 26\]

  5. Subtract 2x from both sides.

    \[7 x - 17 = 26\]

  6. Add 17 to both sides.

    \[7 x = 43\]

  7. Divide both sides by 7 and keep the exact fraction.

    \[x = \frac{43}{7}\]

Check in the original equation

Substitute \(x = \frac{43}{7}\) and evaluate each side separately.

Left side

\[\begin{aligned}&\dfrac{2[3(\left(\frac{43}{7}\right) - 1) + 4]}{5} - \dfrac{3(\left(\frac{43}{7}\right) + 2) - 1}{4} \\&= \frac{67}{35}\end{aligned}\]

Right side

\[\dfrac{\left(\frac{43}{7}\right) + 13}{10} = \frac{67}{35}\]

Both sides equal \(\frac{67}{35}\), so \(x = \frac{43}{7}\) is correct.

Common questions

Why must I do the same thing to both sides?

An equation is a statement of equality. Adding or subtracting the same quantity on both sides, or multiplying or dividing both sides by the same nonzero constant, preserves that equality.

Why does a sign change when a term moves?

The term is not really moving by itself. Subtracting 5 from both sides removes a positive 5 from one side and introduces a subtraction of 5 on the other. Write the operation explicitly when you are unsure.

Can a correct answer be negative or a fraction?

Yes. The value only needs to make the original equation true. Keep fractional answers exact; a rounded decimal may fail an exact substitution check.

How do I avoid mistakes with a negative sign before parentheses?

Treat the sign and number as one factor. Multiply that factor by every term in the parentheses, including each sign. A negative factor times a negative term gives a positive term.

Is this lesson part of a numbered course?

No. This is a standalone lesson with 50 progressively more involved equations, detailed hidden solutions, and a check for every answer. No course enrollment or earlier numbered lesson is required.

A final habit: explain the operation

For each line, name what you did to both sides. For each final answer, substitute it into the original equation. This turns an answer into a solution you can justify.

Return to the first question or explore the SOMATH Journal for more independent learning resources.