Solve 3 * 2 7/15 - 5 1/2 * 7/9 + 1 7/48 * 2 2/11 - 2/5 * 6 5/9 | 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

$$3\cdot2\frac{7}{15}-5\frac{1}{2}\cdot\frac{7}{9}+1\frac{7}{48}\cdot2\frac{2}{11}-\frac{2}{5}\cdot6\frac{5}{9}$$

Evaluate

$3$

Solution Steps

Multiply $2$ and $15$ to get $30$.

$$3\times \frac{30+7}{15}-\frac{5\times 2+1}{2}\times \frac{7}{9}+\frac{1\times 48+7}{48}\times \frac{2\times 11+2}{11}-\frac{2}{5}\times \frac{6\times 9+5}{9}$$

Add $30$ and $7$ to get $37$.

$$3\times \frac{37}{15}-\frac{5\times 2+1}{2}\times \frac{7}{9}+\frac{1\times 48+7}{48}\times \frac{2\times 11+2}{11}-\frac{2}{5}\times \frac{6\times 9+5}{9}$$

Express $3\times \frac{37}{15}$ as a single fraction.

$$\frac{3\times 37}{15}-\frac{5\times 2+1}{2}\times \frac{7}{9}+\frac{1\times 48+7}{48}\times \frac{2\times 11+2}{11}-\frac{2}{5}\times \frac{6\times 9+5}{9}$$

Multiply $3$ and $37$ to get $111$.

$$\frac{111}{15}-\frac{5\times 2+1}{2}\times \frac{7}{9}+\frac{1\times 48+7}{48}\times \frac{2\times 11+2}{11}-\frac{2}{5}\times \frac{6\times 9+5}{9}$$

Reduce the fraction $\frac{111}{15}$ to lowest terms by extracting and canceling out $3$.

$$\frac{37}{5}-\frac{5\times 2+1}{2}\times \frac{7}{9}+\frac{1\times 48+7}{48}\times \frac{2\times 11+2}{11}-\frac{2}{5}\times \frac{6\times 9+5}{9}$$

Multiply $5$ and $2$ to get $10$.

$$\frac{37}{5}-\frac{10+1}{2}\times \frac{7}{9}+\frac{1\times 48+7}{48}\times \frac{2\times 11+2}{11}-\frac{2}{5}\times \frac{6\times 9+5}{9}$$

Add $10$ and $1$ to get $11$.

$$\frac{37}{5}-\frac{11}{2}\times \frac{7}{9}+\frac{1\times 48+7}{48}\times \frac{2\times 11+2}{11}-\frac{2}{5}\times \frac{6\times 9+5}{9}$$

Multiply $\frac{11}{2}$ times $\frac{7}{9}$ by multiplying numerator times numerator and denominator times denominator.

$$\frac{37}{5}-\frac{11\times 7}{2\times 9}+\frac{1\times 48+7}{48}\times \frac{2\times 11+2}{11}-\frac{2}{5}\times \frac{6\times 9+5}{9}$$

Do the multiplications in the fraction $\frac{11\times 7}{2\times 9}$.

$$\frac{37}{5}-\frac{77}{18}+\frac{1\times 48+7}{48}\times \frac{2\times 11+2}{11}-\frac{2}{5}\times \frac{6\times 9+5}{9}$$

Least common multiple of $5$ and $18$ is $90$. Convert $\frac{37}{5}$ and $\frac{77}{18}$ to fractions with denominator $90$.

$$\frac{666}{90}-\frac{385}{90}+\frac{1\times 48+7}{48}\times \frac{2\times 11+2}{11}-\frac{2}{5}\times \frac{6\times 9+5}{9}$$

Since $\frac{666}{90}$ and $\frac{385}{90}$ have the same denominator, subtract them by subtracting their numerators.

$$\frac{666-385}{90}+\frac{1\times 48+7}{48}\times \frac{2\times 11+2}{11}-\frac{2}{5}\times \frac{6\times 9+5}{9}$$

Subtract $385$ from $666$ to get $281$.

$$\frac{281}{90}+\frac{1\times 48+7}{48}\times \frac{2\times 11+2}{11}-\frac{2}{5}\times \frac{6\times 9+5}{9}$$

Multiply $1$ and $48$ to get $48$.

$$\frac{281}{90}+\frac{48+7}{48}\times \frac{2\times 11+2}{11}-\frac{2}{5}\times \frac{6\times 9+5}{9}$$

Add $48$ and $7$ to get $55$.

$$\frac{281}{90}+\frac{55}{48}\times \frac{2\times 11+2}{11}-\frac{2}{5}\times \frac{6\times 9+5}{9}$$

Multiply $2$ and $11$ to get $22$.

$$\frac{281}{90}+\frac{55}{48}\times \frac{22+2}{11}-\frac{2}{5}\times \frac{6\times 9+5}{9}$$

Add $22$ and $2$ to get $24$.

$$\frac{281}{90}+\frac{55}{48}\times \frac{24}{11}-\frac{2}{5}\times \frac{6\times 9+5}{9}$$

Multiply $\frac{55}{48}$ times $\frac{24}{11}$ by multiplying numerator times numerator and denominator times denominator.

$$\frac{281}{90}+\frac{55\times 24}{48\times 11}-\frac{2}{5}\times \frac{6\times 9+5}{9}$$

Do the multiplications in the fraction $\frac{55\times 24}{48\times 11}$.

$$\frac{281}{90}+\frac{1320}{528}-\frac{2}{5}\times \frac{6\times 9+5}{9}$$

Reduce the fraction $\frac{1320}{528}$ to lowest terms by extracting and canceling out $264$.

$$\frac{281}{90}+\frac{5}{2}-\frac{2}{5}\times \frac{6\times 9+5}{9}$$

Least common multiple of $90$ and $2$ is $90$. Convert $\frac{281}{90}$ and $\frac{5}{2}$ to fractions with denominator $90$.

$$\frac{281}{90}+\frac{225}{90}-\frac{2}{5}\times \frac{6\times 9+5}{9}$$

Since $\frac{281}{90}$ and $\frac{225}{90}$ have the same denominator, add them by adding their numerators.

$$\frac{281+225}{90}-\frac{2}{5}\times \frac{6\times 9+5}{9}$$

Add $281$ and $225$ to get $506$.

$$\frac{506}{90}-\frac{2}{5}\times \frac{6\times 9+5}{9}$$

Reduce the fraction $\frac{506}{90}$ to lowest terms by extracting and canceling out $2$.

$$\frac{253}{45}-\frac{2}{5}\times \frac{6\times 9+5}{9}$$

Multiply $6$ and $9$ to get $54$.

$$\frac{253}{45}-\frac{2}{5}\times \frac{54+5}{9}$$

Add $54$ and $5$ to get $59$.

$$\frac{253}{45}-\frac{2}{5}\times \frac{59}{9}$$

Multiply $\frac{2}{5}$ times $\frac{59}{9}$ by multiplying numerator times numerator and denominator times denominator.

$$\frac{253}{45}-\frac{2\times 59}{5\times 9}$$

Do the multiplications in the fraction $\frac{2\times 59}{5\times 9}$.

$$\frac{253}{45}-\frac{118}{45}$$

Since $\frac{253}{45}$ and $\frac{118}{45}$ have the same denominator, subtract them by subtracting their numerators.

$$\frac{253-118}{45}$$

Subtract $118$ from $253$ to get $135$.

$$\frac{135}{45}$$

Divide $135$ by $45$ to get $3$.

$$3$$

Factor

$3$