Solve (((B)/(E))+A)-R=((((331+337+347)/(sqrt((15)^(2)+(10)^(3))))+(25+(18+11)))-(2\times30) | 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

$$(( \frac{ B }{ E } )+A)-R=((( \frac{ 331+337+347 }{ \sqrt{ { 15 }^{ 2 } + { 10 }^{ 3 } } } )+(25+(18+11)))-(2 \times 30)$$

Solve for A

$A=R-\frac{B}{E}+23$
$E\neq 0$

Steps for Solving Linear Equation

Multiply both sides of the equation by $E$.

$$B+EA-RE=E\left(\frac{331+337+347}{\sqrt{15^{2}+10^{3}}}+25+18+11\right)-E\left(2\times 30\right)$$

Add $331$ and $337$ to get $668$.

$$B+EA-RE=E\left(\frac{668+347}{\sqrt{15^{2}+10^{3}}}+25+18+11\right)-E\left(2\times 30\right)$$

Add $668$ and $347$ to get $1015$.

$$B+EA-RE=E\left(\frac{1015}{\sqrt{15^{2}+10^{3}}}+25+18+11\right)-E\left(2\times 30\right)$$

Calculate $15$ to the power of $2$ and get $225$.

$$B+EA-RE=E\left(\frac{1015}{\sqrt{225+10^{3}}}+25+18+11\right)-E\left(2\times 30\right)$$

Calculate $10$ to the power of $3$ and get $1000$.

$$B+EA-RE=E\left(\frac{1015}{\sqrt{225+1000}}+25+18+11\right)-E\left(2\times 30\right)$$

Add $225$ and $1000$ to get $1225$.

$$B+EA-RE=E\left(\frac{1015}{\sqrt{1225}}+25+18+11\right)-E\left(2\times 30\right)$$

Calculate the square root of $1225$ and get $35$.

$$B+EA-RE=E\left(\frac{1015}{35}+25+18+11\right)-E\left(2\times 30\right)$$

Divide $1015$ by $35$ to get $29$.

$$B+EA-RE=E\left(29+25+18+11\right)-E\left(2\times 30\right)$$

Add $29$ and $25$ to get $54$.

$$B+EA-RE=E\left(54+18+11\right)-E\left(2\times 30\right)$$

Add $54$ and $18$ to get $72$.

$$B+EA-RE=E\left(72+11\right)-E\left(2\times 30\right)$$

Add $72$ and $11$ to get $83$.

$$B+EA-RE=E\times 83-E\left(2\times 30\right)$$

Subtract $B$ from both sides.

$$EA-RE=E\times 83-E\left(2\times 30\right)-B$$

Add $RE$ to both sides.

$$EA=E\times 83-E\left(2\times 30\right)-B+RE$$

Reorder the terms.

$$AE=-E\left(2\times 30\right)+ER-B+83E$$

The equation is in standard form.

$$EA=23E+ER-B$$

Divide both sides by $E$.

$$\frac{EA}{E}=\frac{23E+ER-B}{E}$$

Dividing by $E$ undoes the multiplication by $E$.

$$A=\frac{23E+ER-B}{E}$$

Divide $23E-B+ER$ by $E$.

$$A=R-\frac{B}{E}+23$$

Solve for B

$B=-E\left(A-R-23\right)$
$E\neq 0$