Solve ((B)/(E)+A)-R=((227+229+233+239)/((2)^(5))+(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

$$( \frac{ B }{ E } +A)-R=( \frac{ 227+229+233+239 }{ { 2 }^{ 5 } } +(25+(18+11))-(2 \times 30)$$

Solve for A

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

Steps for Solving Linear Equation

Multiply both sides of the equation by $E$.

$$B+EA-RE=E\left(\frac{227+229+233+239}{2^{5}}+25+18+11\right)-E\left(2\times 30\right)$$

Add $227$ and $229$ to get $456$.

$$B+EA-RE=E\left(\frac{456+233+239}{2^{5}}+25+18+11\right)-E\left(2\times 30\right)$$

Add $456$ and $233$ to get $689$.

$$B+EA-RE=E\left(\frac{689+239}{2^{5}}+25+18+11\right)-E\left(2\times 30\right)$$

Add $689$ and $239$ to get $928$.

$$B+EA-RE=E\left(\frac{928}{2^{5}}+25+18+11\right)-E\left(2\times 30\right)$$

Calculate $2$ to the power of $5$ and get $32$.

$$B+EA-RE=E\left(\frac{928}{32}+25+18+11\right)-E\left(2\times 30\right)$$

Divide $928$ by $32$ 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

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