Linear Equation Problems

A number that turns an equation into a true equality is called a root, or solution, of the equation. Two equations are equivalent if they have exactly the same roots. Each of these moves turns an equation into an equivalent one:

1. replacing an expression with one that is identically equal to it, for example ${2(x+3)}$ with ${2x+6}$;
2. moving a term from one side to the other and changing its sign;
3. multiplying or dividing both sides by the same number, as long as it is not zero.

An equation that can be brought to the form ${ax+b=0}$, where $a$ and $b$ are numbers, is a linear equation in $x$. The lesson Solving Linear Equations explains these moves with examples. Below are problems with full solutions.

Problem 1

Solve the equation:

A) ${16x+10-32}={35-10x-5}$
B) ${y+\frac{3}{2}y+25}={\frac{1}{2}y+\frac{3}{4}y-\frac{5}{2}y+y+37}$
C) ${7u-9-3u+5}={11u-6-4u}$

Solution:

A) Simplify both sides, then move the terms:

${{16x-22}={30-10x}}$

${{16x+10x}={30+22}}$, so ${26x=52}$ and ${x=2}$

B) Collect the $y$ terms on each side. On the left ${1+\frac{3}{2}=\frac{5}{2}}$, and on the right ${\frac{1}{2}+\frac{3}{4}-\frac{5}{2}+1=-\frac{1}{4}}$:

${{\frac{5}{2}y+25}={-\frac{1}{4}y+37}}$

${{\frac{5}{2}y+\frac{1}{4}y}={37-25}}$, so ${\frac{11}{4}y=12}$ and ${y=\frac{48}{11}}$

C)

${{4u-4}={7u-6}}$

${{6-4}={7u-4u}}$, so ${2=3u}$ and ${u=\frac{2}{3}}$

Problem 2

Solve the equation:

A) ${7(3x-6)+5(x-3)}-{2(x-7)}={5}$
B) ${(x-3)(x+4)-2(3x-2)}={(x-4)^2}$
C) ${(x+1)^3-(x-1)^3}={6(x^2+x+1)}$

Solution:

A)

${21x-42+5x-15-2x+14}={5}$

${21x+5x-2x}={5+42+15-14}$, so ${24x=48}$ and ${x=2}$

B)

${x^2+4x-3x-12-6x+4}={x^2-8x+16}$

${x^2-5x-8}={x^2-8x+16}$

The ${x^2}$ terms cancel:

${{8x-5x}={16+8}}$, so ${3x=24}$ and ${x=8}$

C)

${x^3+3x^2+3x+1}-{(x^3-3x^2+3x-1)}={6x^2+6x+6}$

${{6x^2+2}={6x^2+6x+6}}$

${{2}={6x+6}}$, so ${6x=-4}$ and ${x=-\frac{2}{3}}$

Problem 3

Solve the equation:

A) ${{\frac{5x-4}{2}}={\frac{0.5x+1}{3}}}$
B) ${{1-\frac{x-3}{5}}={\frac{-3x+3}{3}}}$
C) ${{\frac{x+1}{3}-\frac{2x+5}{2}}={-3}}$
D) ${\frac{3(x-1)}{2}+\frac{2(x+2)}{4}}={\frac{3x+4.5}{5}}$

Solution:

A) Cross-multiply:

${{3(5x-4)}={2(0.5x+1)}}$

${{15x-12}={x+2}}$, so ${14x=14}$ and ${x=1}$

B) The right side is ${\frac{3(1-x)}{3}=1-x}$, so

${{1-\frac{x-3}{5}}={1-x}}$

Subtract 1 from both sides and multiply by $-5$:

${{x-3}={5x}}$, so ${-3=4x}$ and ${x=-\frac{3}{4}}$

C) Multiply by 6:

${2(x+1)-3(2x+5)}={-18}$

${{2x+2-6x-15}={-18}}$

${{-4x-13}={-18}}$, so ${-4x=-5}$ and ${x=\frac{5}{4}}$

D) The common denominator of 2, 4 and 5 is 20. Multiply by 20:

${30(x-1)+10(x+2)}={4(3x+4.5)}$

${30x-30+10x+20}={12x+18}$

${{40x-12x}={18+10}}$, so ${28x=28}$ and ${x=1}$

Problem 4

Prove that every number is a root of the equation:

A) ${7x-13=-13+7x}$
B) ${\left(\frac{1}{2}-x\right)^2-\left(\frac{1}{2}+x\right)^2}={-2x}$
C) ${{3x-3x}={26-2(7+6)}}$
D) ${{\frac{-3x+4x^2}{5}}={(0.8x-0.6)x}}$

Solution: An equation holds for every number if it can be brought to ${0\cdot x=0}$, or if its two sides become the same expression. Then any value of $x$ gives a true equality.

A) ${{7x-7x}={-13+13}}$, so ${0\cdot x=0}$, and every number is a root.

B) Expand the squares:

${\frac{1}{4}-x+x^2-\left(\frac{1}{4}+x+x^2\right)}={-2x}$

${{-2x}={-2x}}$

Both sides are the same, so every number is a root.

C) ${{0\cdot x}={26-2\cdot 13}}$, so ${0\cdot x=0}$, and every number is a root.

D) Multiply by 5:

${{-3x+4x^2}={5(0.8x-0.6)x}}$

${{-3x+4x^2}={4x^2-3x}}$

Both sides are the same, so every number is a root.

Problem 5

Prove that the equation has no roots:

A) ${0\cdot x=34}$
B) ${5-3x=7-3x}$
C) ${{\frac{x-3}{4}}={\frac{x+5}{4}}}$
D) ${2(3x-1)-3(2x+1)}={6}$

Solution: An equation that can be brought to ${0\cdot x=c}$ with ${c\ne 0}$ has no roots: the left side is 0 for every $x$, and the right side is not.

A) The left side is 0 for every $x$, and the right side is 34. No number makes the equality true.

B) ${{-3x+3x}={7-5}}$, so ${0\cdot x=2}$: no roots.

C) Multiply by 4: ${{x-3}={x+5}}$, so ${{x-x}={5+3}}$ and ${0\cdot x=8}$: no roots.

D) ${{6x-2-6x-3}={6}}$, so ${0\cdot x=11}$: no roots.

Problem 6

Solve the equation:

A) ${2x^2-3(1-x)(x+2)}+{(x-4)(1-5x)+58}={0}$
B) ${3(x+1)^2-(3x+5)x}={x+3}$
C) ${{x^2-(x-1)(x+1)}={4}}$
D) ${(x-1)(x^2+x+1)}={(x-1)^3+3x(x-1)}$
E) ${(3x-1)^2-x(15x+7)}={x(x+1)(x-1)-(x+2)^3}$

Solution: These equations contain squares and cubes, but once the brackets are multiplied out, the ${x^2}$ and ${x^3}$ terms cancel and a linear equation is left.

A)

${2x^2-3(x+2-x^2-2x)}+{x-5x^2-4+20x+58}={0}$

${2x^2-3x-6+3x^2+6x}+{x-5x^2-4+20x+58}={0}$

${{24x+48}={0}}$, so ${24x=-48}$ and ${x=-2}$

B)

${3x^2+6x+3-3x^2-5x}={x+3}$

${{x+3}={x+3}}$

Both sides are the same, so every number is a root.

C)

${{x^2-(x^2-1)}={4}}$, so ${0\cdot x=3}$: no roots.

D)

${x^3+x^2+x-x^2-x-1}={x^3-3x^2+3x-1+3x^2-3x}$

${{x^3-1}={x^3-1}}$

Both sides are the same, so every number is a root.

E)

${9x^2-6x+1-15x^2-7x}={x^3-x-(x^3+6x^2+12x+8)}$

${-6x^2-13x+1}={-6x^2-13x-8}$

${0\cdot x=-9}$: no roots.

Problem 7

Solve the equation:

A) ${\frac{6x-1}{5}-\frac{1-2x}{2}}={\frac{12x+49}{10}}$
B) ${\frac{x-3}{2}+\frac{2x-2}{4}}={\frac{7x-6}{3}}$

Solution:

A) Multiply by 10:

${2(6x-1)-5(1-2x)}={12x+49}$

${12x-2-5+10x}={12x+49}$

${{10x}={56}}$, so ${x=5.6}$

B) Since ${\frac{2x-2}{4}=\frac{x-1}{2}}$, the left side is ${\frac{x-3+x-1}{2}=\frac{2x-4}{2}}$:

${{\frac{2x-4}{2}}={\frac{7x-6}{3}}}$

Cross-multiply:

${{3(2x-4)}={2(7x-6)}}$

${{6x-12}={14x-12}}$, so ${0=8x}$ and ${x=0}$

Problem 8

The function ${f(x)=x+4}$ is given. Solve the equation

${{\frac{3f(x-2)}{f(0)}+4}={f(2x+1)}}$

Solution: First find the values of $f$ that appear in the equation:

${f(0)=0+4=4}$, ${f(x-2)=x-2+4=x+2}$, ${f(2x+1)=2x+1+4=2x+5}$

The equation becomes

${{\frac{3(x+2)}{4}+4}={2x+5}}$

Multiply by 4:

${3(x+2)+16}={4(2x+5)}$

${{3x+22}={8x+20}}$, so ${2=5x}$ and ${x=0.4}$

Problem 9

Solve the equation

${(2x-1)^2-x(10x+1)}={x(1-x)(1+x)-(2-x)^3}$

Solution: Simplify each side:

${4x^2-4x+1-10x^2-x}={x-x^3-(8-12x+6x^2-x^3)}$

${-6x^2-5x+1}={-6x^2+13x-8}$

${{9}={18x}}$, so ${x=\frac{1}{2}}$

Problem 10

Solve the equation

${(2x+3)^2-x(1+2x)(1-2x)}={(2x-1)^2+4x^3-1}$

Solution:

${4x^2+12x+9-x(1-4x^2)}={4x^2-4x+1+4x^3-1}$

${4x^3+4x^2+11x+9}={4x^3+4x^2-4x}$

${{11x+9}={-4x}}$, so ${15x=-9}$ and ${x=-\frac{3}{5}}$

Problem 11

Solve the equation

${(2x-1)^3+2x(2x-3)(3-2x)}-{(3x-1)^2}={3x^2-2}$

Solution: Since ${3-2x=-(2x-3)}$, the middle term is ${-2x(2x-3)^2}$. Use the formulas for ${(a-b)^3}$ and ${(a-b)^2}$:

${8x^3-12x^2+6x-1}-{2x(4x^2-12x+9)}-{(9x^2-6x+1)}={3x^2-2}$

${8x^3-12x^2+6x-1}-{8x^3}+{24x^2-18x}-{9x^2}+{6x-1}={3x^2-2}$

${{3x^2-6x-2}={3x^2-2}}$, so ${-6x=0}$ and ${x=0}$

Problem 12

Solve the equation

${\left(2x-\frac{1}{2}\right)^2-(2x-3)(2x+3)}={x+\frac{1}{4}}$

Solution:

${4x^2-2x+\frac{1}{4}-(4x^2-9)}={x+\frac{1}{4}}$

${{-2x+\frac{1}{4}+9}={x+\frac{1}{4}}}$

${{9}={3x}}$, so ${x=3}$

Problem 13

Prove that the two equations are equivalent:

A) ${\frac{x-5}{2}+\frac{x-1}{8}}={\frac{1.5x-10}{4}}$ and ${\frac{x+6}{2}-\frac{5.5-0.5x}{3}}={1.5}$
B) ${x-\frac{8x+7}{6}+\frac{x}{3}}={-\frac{7}{6}}$ and ${{2x-\frac{6-x}{3}-\frac{7}{3}x}={-2}}$

Solution: Two equations are equivalent if they have the same roots, so solve each of them.

A) Multiply the first equation by 8:

${4(x-5)+x-1}={2(1.5x-10)}$

${{4x-20+x-1}={3x-20}}$

${{5x-3x}={-20+21}}$, so ${2x=1}$ and ${x=\frac{1}{2}}$

Multiply the second equation by 6:

${3(x+6)-2(5.5-0.5x)}={9}$

${{3x+18-11+x}={9}}$, so ${4x=2}$ and ${x=\frac{1}{2}}$

Both equations have the single root ${\frac{1}{2}}$, so they are equivalent.

B) Multiply the first equation by 6:

${{6x-(8x+7)+2x}={-7}}$, so ${-7=-7}$

Multiply the second equation by 3:

${{6x-(6-x)-7x}={-6}}$, so ${-6=-6}$

Both equalities are true whatever $x$ is. Every number is a root of both equations, so they are equivalent.

Problem 14

Solve the equation:

A) ${(2x+1)^2-x(1-2x)(1+2x)}={(2x-1)^2+4x^3-3}$
B) ${(2x-1)^2+(x-2)^3}={x^2(x-2)+8x-7}$
C) ${(x+2)(x^2-2x+4)}+{x(1-x)(1+x)}={x-4}$
D) ${\frac{8x+5}{4}-\frac{1}{2}\left[2-\frac{3-x}{3}\right]}={2x+\frac{5}{6}}$
E) ${\frac{x}{3}-\frac{x+3}{4}}={x-\frac{1}{3}\left[1-\frac{3-24x}{8}\right]}$
F) ${\frac{x}{5}-\frac{(2x-3)^2}{3}}={\frac{1}{5}\left[5-\frac{20x^2-43x}{3}\right]}$

Solution:

A)

${4x^2+4x+1-x(1-4x^2)}={4x^2-4x+1+4x^3-3}$

${4x^3+4x^2+3x+1}={4x^3+4x^2-4x-2}$

${{3x+1}={-4x-2}}$, so ${7x=-3}$ and ${x=-\frac{3}{7}}$

B)

${4x^2-4x+1}+{x^3-6x^2+12x-8}={x^3-2x^2+8x-7}$

${x^3-2x^2+8x-7}={x^3-2x^2+8x-7}$

Both sides are the same, so every number is a root.

C)

${{x^3+8+x-x^3}={x-4}}$

${{x+8}={x-4}}$, so ${0\cdot x=-12}$: no roots.

D) Open the square brackets:

${\frac{8x+5}{4}-1+\frac{3-x}{6}}={2x+\frac{5}{6}}$

Multiply by 12:

${3(8x+5)-12+2(3-x)}={24x+10}$

${24x+15-12+6-2x}={24x+10}$

${{22x+9}={24x+10}}$, so ${-1=2x}$ and ${x=-\frac{1}{2}}$

E) Open the square brackets:

${\frac{x}{3}-\frac{x+3}{4}}={x-\frac{1}{3}+\frac{3-24x}{24}}$

Multiply by 24:

${8x-6(x+3)}={24x-8+3-24x}$

${{2x-18}={-5}}$, so ${2x=13}$ and ${x=6.5}$

F) Open the square brackets:

${\frac{x}{5}-\frac{(2x-3)^2}{3}}={1-\frac{20x^2-43x}{15}}$

Multiply by 15:

${3x-5(4x^2-12x+9)}={15-(20x^2-43x)}$

${3x-20x^2+60x-45}={15-20x^2+43x}$

${{63x-45}={15+43x}}$, so ${20x=60}$ and ${x=3}$

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