Solving Linear Equations

Solve $5x-11=2x+7$, and then write its solution set.

$$5x-11=2x+7$$

The goal is eventually to reach an equation of the form $x=\text{constant}$ (like $x=2$) or $\text{constant}=x$ (like $2=x$).

On the left side, the term $-11$ is not wanted.

$$5x-11=2x+7$$

Adding 11 to the left side will make that term go away. Use the addition property of equality. Add 11 to each side.

$$5x-11+11=2x+7+11$$

Simplify the sides.

$$5x=2x+18$$

On the right side, the term $2x$ is not wanted.

$$5x=2x+18$$

Use the addition property of equality again. Subtract $2x$ from each side.

$$5x-2x=2x+18-2x$$

Simplify the sides.

$$3x=18$$

The $3$ in the left side $3x$ is not wanted.

$$3x=18$$

Dividing the left side by 3 will make that $3$ go away. Use the multiplication property of equality. Divide each side by 3.

$$\frac{3x}{3}=\frac{18}{3}$$

Simplify the sides.

$$x=6$$

$$\left\{6\right\}$$

Solve $5x+1=6-3\left(x-5\right)$, and then write its solution set.

$$5x+1=6-3\left(x-5\right)$$

Simplify the right side until it is in simplest form. The left side is already in simplest form, so it must patiently wait for the right side to be simplified.

$$5x+1=6-3x+15$$

$$5x+1=21-3x$$

Use the addition property of equality.

$$5x+1-1=21-3x-1$$

Simplify the sides.

$$5x=20-3x$$

Use the addition property of equality.

$$5x+3x=20-3x+3x$$

Simplify the sides.

$$8x=20$$

Use the multiplication property of equality.

$$\frac{8x}{8}=\frac{20}{8}$$

Simplify the sides.

$$x=\frac{20}{8}$$

Simplify the right side to lowest terms.

$$x=\frac{4·5}{4·2}$$

$$x=\frac{5}{2}$$

$$\left\{\frac{5}{2}\right\}$$

Solve $12x+5=-7x-90$, and then write its solution set.

$$12x+5=-7x-90$$

$$12x+5-5=-7x-90-5$$

$$12x=-7x-95$$

$$12x+7x=-7x-95+7x$$

$$19x=-95$$

$$\frac{19x}{19}=\frac{-95}{19}$$

$$x=-5$$

$$\left\{-5\right\}$$

Solve $6\left(2+5x\right)+4\left(3x-2\right)=15-\left(x+1\right)-2\left(-5x-3\right)$, and then write its solution set.

$$6\left(2+5x\right)+4\left(3x-2\right)=15-\left(x+1\right)-2\left(-5x-3\right)$$

What a mess. Simplify the sides.

$$6\left(2+5x\right)+4\left(3x-2\right)=15-\left(x+1\right)-2\left(-5x-3\right)$$

$$12+30x+4\left(3x-2\right)=15-x-1-2\left(-5x-3\right)$$

$$12+30x+4\left(3x-2\right)=15-x-1-2\left(-5x-3\right)$$

$$12+30x+12x-8=15-x-1+10x+6$$

$$12+30x+12x-8=15-x-1+10x+6$$

$$12+42x-8=15+9x-1+6$$

$$12+42x-8=15+9x-1+6$$

$$4+42x=20+9x$$

$$4+42x=20+9x$$

Use the addition property of equality.

$$4+42x-4=20+9x-4$$

$$42x=16+9x$$

Use the addition property of equality.

$$42x-9x=16+9x-9x$$

$$33x=16$$

Use the multiplication property of equality.

$$\frac{33x}{33}=\frac{16}{33}$$

Simplify the left side.

$$x=\frac{16}{33}$$

The right side is in lowest terms.

$$\left\{\frac{16}{33}\right\}$$

Solve $4x+1=4x+6$, and then write its solution set.

$$4x+1=4x+6$$

$$4x+1-4x=4x+6-4x$$

$$1=6$$

Since $1=6$ is false, the equation has no solution. The solution set has nothing in it. The special symbol $\varnothing$ is used to write the empty set, which is the set that has nothing in it.

Do not write $\left\{\right\}$ for the empty set. The empty set has nothing in it. The set $\left\{\right\}$ does have something in it: the empty set.

$$\varnothing$$

Solve $3\left(2x+4\right)-2=5\left(x+2\right)+x$, and then write its solution set.

$$3\left(2x+4\right)-2=5\left(x+2\right)+x$$

$$6x+12-2=5x+10+x$$

$$6x+10=6x+10$$

$$6x+10-6x=6x+10-6x$$

$$10=10$$

Since $10=10$ is true, the equation is true no matter what number is substituted for x. Every real number is in the equation's solution set.

$$\left\{\text{all real numbers}\right\}$$

Solve $\frac{2}{5}x+\frac{3}{2}=\frac{1}{2}x+\frac{3}{5}$, and then write its solution set.

$$\frac{2}{5}x+\frac{3}{2}=\frac{1}{2}x+\frac{3}{5}$$

The least common denominator ( LCD) of the fractions in the equation is 10. Use the multiplication property of equality. Multiply each side by 10.

$$10\left(\frac{2}{5}x+\frac{3}{2}\right)=10\left(\frac{1}{2}x+\frac{3}{5}\right)$$

$$10\left(\frac{2}{5}x\right)+10\left(\frac{3}{2}\right)=10\left(\frac{1}{2}x\right)+10\left(\frac{3}{5}\right)$$

$$\frac{10·2}{5}x+\frac{10·3}{2}=\frac{10·1}{2}x+\frac{10·3}{5}$$

$$\frac{5·2·2}{5}x+\frac{2·5·3}{2}=\frac{2·5·1}{2}x+\frac{5·2·3}{5}$$

$$\frac{5·2·2}{5}x+\frac{2·5·3}{2}=\frac{2·5·1}{2}x+\frac{5·2·3}{5}$$

$$4x+15=5x+6$$

$$4x+15=5x+6$$

$$4x+15-4x=5x+6-4x$$

$$15=x+6$$

$$15-6=x+6-6$$

$$9=x$$

$$\left\{9\right\}$$