Solve (1)/(3)(x-3)+(2)/(3)=(1)/(3)(4x-3)+(7)/(2) | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$\frac{ 1 }{ 3 } (x-3)+ \frac{ 2 }{ 3 } = \frac{ 1 }{ 3 } (4x-3)+ \frac{ 7 }{ 2 }$$

Solve for x

$x = -\frac{17}{6} = -2\frac{5}{6} \approx -2.833333333$

Steps for Solving Linear Equation

Use the distributive property to multiply $\frac{1}{3}$ by $x-3$.

$$\frac{1}{3}x+\frac{1}{3}\left(-3\right)+\frac{2}{3}=\frac{1}{3}\left(4x-3\right)+\frac{7}{2}$$

Multiply $\frac{1}{3}$ and $-3$ to get $\frac{-3}{3}$.

$$\frac{1}{3}x+\frac{-3}{3}+\frac{2}{3}=\frac{1}{3}\left(4x-3\right)+\frac{7}{2}$$

Divide $-3$ by $3$ to get $-1$.

$$\frac{1}{3}x-1+\frac{2}{3}=\frac{1}{3}\left(4x-3\right)+\frac{7}{2}$$

Convert $-1$ to fraction $-\frac{3}{3}$.

$$\frac{1}{3}x-\frac{3}{3}+\frac{2}{3}=\frac{1}{3}\left(4x-3\right)+\frac{7}{2}$$

Since $-\frac{3}{3}$ and $\frac{2}{3}$ have the same denominator, add them by adding their numerators.

$$\frac{1}{3}x+\frac{-3+2}{3}=\frac{1}{3}\left(4x-3\right)+\frac{7}{2}$$

Add $-3$ and $2$ to get $-1$.

$$\frac{1}{3}x-\frac{1}{3}=\frac{1}{3}\left(4x-3\right)+\frac{7}{2}$$

Use the distributive property to multiply $\frac{1}{3}$ by $4x-3$.

$$\frac{1}{3}x-\frac{1}{3}=\frac{1}{3}\times 4x+\frac{1}{3}\left(-3\right)+\frac{7}{2}$$

Multiply $\frac{1}{3}$ and $4$ to get $\frac{4}{3}$.

$$\frac{1}{3}x-\frac{1}{3}=\frac{4}{3}x+\frac{1}{3}\left(-3\right)+\frac{7}{2}$$

Multiply $\frac{1}{3}$ and $-3$ to get $\frac{-3}{3}$.

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

Divide $-3$ by $3$ to get $-1$.

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

Convert $-1$ to fraction $-\frac{2}{2}$.

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

Since $-\frac{2}{2}$ and $\frac{7}{2}$ have the same denominator, add them by adding their numerators.

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

Add $-2$ and $7$ to get $5$.

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

Subtract $\frac{4}{3}x$ from both sides.

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

Combine $\frac{1}{3}x$ and $-\frac{4}{3}x$ to get $-x$.

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

Add $\frac{1}{3}$ to both sides.

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

Least common multiple of $2$ and $3$ is $6$. Convert $\frac{5}{2}$ and $\frac{1}{3}$ to fractions with denominator $6$.

$$-x=\frac{15}{6}+\frac{2}{6}$$

Since $\frac{15}{6}$ and $\frac{2}{6}$ have the same denominator, add them by adding their numerators.

$$-x=\frac{15+2}{6}$$

Add $15$ and $2$ to get $17$.

$$-x=\frac{17}{6}$$

Multiply both sides by $-1$.

$$x=-\frac{17}{6}$$