Solve (A+B)/(C)=(((2\times26)+(1+33+2+1+3+5))+((0\times8)+7))/(((0\times26)+11)) | 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

$$\frac{ A+B }{ C } = \frac{ ((2 \times 26)+(1+33+2+1+3+5))+((0 \times 8)+7) }{ ((0 \times 26)+11) }$$

Solve for A

$A=\frac{104C}{11}-B$
$C\neq 0$

Solution Steps

Multiply both sides of the equation by $C$.

$$A+B=\frac{1}{11}C\left(2\times 26+1+33+2+1+3+5+0\times 8+7\right)$$

Do the multiplications.

$$A+B=\frac{1}{11}C\left(52+1+33+2+1+3+5+0+7\right)$$

Add $52$ and $1$ to get $53$.

$$A+B=\frac{1}{11}C\left(53+33+2+1+3+5+0+7\right)$$

Add $53$ and $33$ to get $86$.

$$A+B=\frac{1}{11}C\left(86+2+1+3+5+0+7\right)$$

Add $86$ and $2$ to get $88$.

$$A+B=\frac{1}{11}C\left(88+1+3+5+0+7\right)$$

Add $88$ and $1$ to get $89$.

$$A+B=\frac{1}{11}C\left(89+3+5+0+7\right)$$

Add $89$ and $3$ to get $92$.

$$A+B=\frac{1}{11}C\left(92+5+0+7\right)$$

Add $92$ and $5$ to get $97$.

$$A+B=\frac{1}{11}C\left(97+0+7\right)$$

Add $97$ and $0$ to get $97$.

$$A+B=\frac{1}{11}C\left(97+7\right)$$

Add $97$ and $7$ to get $104$.

$$A+B=\frac{1}{11}C\times 104$$

Multiply $\frac{1}{11}$ and $104$ to get $\frac{104}{11}$.

$$A+B=\frac{104}{11}C$$

Subtract $B$ from both sides.

$$A=\frac{104}{11}C-B$$

Solve for B

$B=\frac{104C}{11}-A$
$C\neq 0$

Solution Steps

Multiply both sides of the equation by $C$.

$$A+B=\frac{1}{11}C\left(2\times 26+1+33+2+1+3+5+0\times 8+7\right)$$

Do the multiplications.

$$A+B=\frac{1}{11}C\left(52+1+33+2+1+3+5+0+7\right)$$

Add $52$ and $1$ to get $53$.

$$A+B=\frac{1}{11}C\left(53+33+2+1+3+5+0+7\right)$$

Add $53$ and $33$ to get $86$.

$$A+B=\frac{1}{11}C\left(86+2+1+3+5+0+7\right)$$

Add $86$ and $2$ to get $88$.

$$A+B=\frac{1}{11}C\left(88+1+3+5+0+7\right)$$

Add $88$ and $1$ to get $89$.

$$A+B=\frac{1}{11}C\left(89+3+5+0+7\right)$$

Add $89$ and $3$ to get $92$.

$$A+B=\frac{1}{11}C\left(92+5+0+7\right)$$

Add $92$ and $5$ to get $97$.

$$A+B=\frac{1}{11}C\left(97+0+7\right)$$

Add $97$ and $0$ to get $97$.

$$A+B=\frac{1}{11}C\left(97+7\right)$$

Add $97$ and $7$ to get $104$.

$$A+B=\frac{1}{11}C\times 104$$

Multiply $\frac{1}{11}$ and $104$ to get $\frac{104}{11}$.

$$A+B=\frac{104}{11}C$$

Subtract $A$ from both sides.

$$B=\frac{104}{11}C-A$$