Solve (5(3-(13)/(18)))/(4)+((13)/(18)+8)/(6)=3((13)/(18)+1) | 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

$$\frac{ 5(3- \frac{ 13 }{ 18 } ) }{ 4 } + \frac{ \frac{ 13 }{ 18 } +8 }{ 6 } =3( \frac{ 13 }{ 18 } +1)$$

Verify

$\text{false}$

Solution Steps

Multiply both sides of the equation by $36$, the least common multiple of $4,6,18$.

$$9\times 5\left(3-\frac{13}{18}\right)+6\left(\frac{13}{18}+8\right)=108\left(\frac{13}{18}+1\right)$$

Multiply $9$ and $5$ to get $45$.

$$45\left(3-\frac{13}{18}\right)+6\left(\frac{13}{18}+8\right)=108\left(\frac{13}{18}+1\right)$$

Convert $3$ to fraction $\frac{54}{18}$.

$$45\left(\frac{54}{18}-\frac{13}{18}\right)+6\left(\frac{13}{18}+8\right)=108\left(\frac{13}{18}+1\right)$$

Since $\frac{54}{18}$ and $\frac{13}{18}$ have the same denominator, subtract them by subtracting their numerators.

$$45\times \frac{54-13}{18}+6\left(\frac{13}{18}+8\right)=108\left(\frac{13}{18}+1\right)$$

Subtract $13$ from $54$ to get $41$.

$$45\times \frac{41}{18}+6\left(\frac{13}{18}+8\right)=108\left(\frac{13}{18}+1\right)$$

Express $45\times \frac{41}{18}$ as a single fraction.

$$\frac{45\times 41}{18}+6\left(\frac{13}{18}+8\right)=108\left(\frac{13}{18}+1\right)$$

Multiply $45$ and $41$ to get $1845$.

$$\frac{1845}{18}+6\left(\frac{13}{18}+8\right)=108\left(\frac{13}{18}+1\right)$$

Reduce the fraction $\frac{1845}{18}$ to lowest terms by extracting and canceling out $9$.

$$\frac{205}{2}+6\left(\frac{13}{18}+8\right)=108\left(\frac{13}{18}+1\right)$$

Convert $8$ to fraction $\frac{144}{18}$.

$$\frac{205}{2}+6\left(\frac{13}{18}+\frac{144}{18}\right)=108\left(\frac{13}{18}+1\right)$$

Since $\frac{13}{18}$ and $\frac{144}{18}$ have the same denominator, add them by adding their numerators.

$$\frac{205}{2}+6\times \frac{13+144}{18}=108\left(\frac{13}{18}+1\right)$$

Add $13$ and $144$ to get $157$.

$$\frac{205}{2}+6\times \frac{157}{18}=108\left(\frac{13}{18}+1\right)$$

Express $6\times \frac{157}{18}$ as a single fraction.

$$\frac{205}{2}+\frac{6\times 157}{18}=108\left(\frac{13}{18}+1\right)$$

Multiply $6$ and $157$ to get $942$.

$$\frac{205}{2}+\frac{942}{18}=108\left(\frac{13}{18}+1\right)$$

Reduce the fraction $\frac{942}{18}$ to lowest terms by extracting and canceling out $6$.

$$\frac{205}{2}+\frac{157}{3}=108\left(\frac{13}{18}+1\right)$$

Least common multiple of $2$ and $3$ is $6$. Convert $\frac{205}{2}$ and $\frac{157}{3}$ to fractions with denominator $6$.

$$\frac{615}{6}+\frac{314}{6}=108\left(\frac{13}{18}+1\right)$$

Since $\frac{615}{6}$ and $\frac{314}{6}$ have the same denominator, add them by adding their numerators.

$$\frac{615+314}{6}=108\left(\frac{13}{18}+1\right)$$

Add $615$ and $314$ to get $929$.

$$\frac{929}{6}=108\left(\frac{13}{18}+1\right)$$

Convert $1$ to fraction $\frac{18}{18}$.

$$\frac{929}{6}=108\left(\frac{13}{18}+\frac{18}{18}\right)$$

Since $\frac{13}{18}$ and $\frac{18}{18}$ have the same denominator, add them by adding their numerators.

$$\frac{929}{6}=108\times \frac{13+18}{18}$$

Add $13$ and $18$ to get $31$.

$$\frac{929}{6}=108\times \frac{31}{18}$$

Express $108\times \frac{31}{18}$ as a single fraction.

$$\frac{929}{6}=\frac{108\times 31}{18}$$

Multiply $108$ and $31$ to get $3348$.

$$\frac{929}{6}=\frac{3348}{18}$$

Divide $3348$ by $18$ to get $186$.

$$\frac{929}{6}=186$$

Convert $186$ to fraction $\frac{1116}{6}$.

$$\frac{929}{6}=\frac{1116}{6}$$

Compare $\frac{929}{6}$ and $\frac{1116}{6}$.

$$\text{false}$$