Solve 5*2.5+2*3.75+1.5*1.5 2.5+1.5 =f4arll | 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

$$\frac{2.5\times2.5+2\times3.75+1.5\times1.5}{2.5+1.5}=f4arll$$

Solve for a

$a=\frac{4}{f_{4}rl^{2}}$
$l\neq 0\text{ and }r\neq 0\text{ and }f_{4}\neq 0$

Steps for Solving Linear Equation

Multiply $l$ and $l$ to get $l^{2}$.

$$\frac{2.5\times 2.5+2\times 3.75+1.5\times 1.5}{2.5+1.5}=f_{4}arl^{2}$$

Multiply $2.5$ and $2.5$ to get $6.25$.

$$\frac{6.25+2\times 3.75+1.5\times 1.5}{2.5+1.5}=f_{4}arl^{2}$$

Multiply $2$ and $3.75$ to get $7.5$.

$$\frac{6.25+7.5+1.5\times 1.5}{2.5+1.5}=f_{4}arl^{2}$$

Add $6.25$ and $7.5$ to get $13.75$.

$$\frac{13.75+1.5\times 1.5}{2.5+1.5}=f_{4}arl^{2}$$

Multiply $1.5$ and $1.5$ to get $2.25$.

$$\frac{13.75+2.25}{2.5+1.5}=f_{4}arl^{2}$$

Add $13.75$ and $2.25$ to get $16$.

$$\frac{16}{2.5+1.5}=f_{4}arl^{2}$$

Add $2.5$ and $1.5$ to get $4$.

$$\frac{16}{4}=f_{4}arl^{2}$$

Divide $16$ by $4$ to get $4$.

$$4=f_{4}arl^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$f_{4}arl^{2}=4$$

The equation is in standard form.

$$f_{4}rl^{2}a=4$$

Divide both sides by $f_{4}rl^{2}$.

$$\frac{f_{4}rl^{2}a}{f_{4}rl^{2}}=\frac{4}{f_{4}rl^{2}}$$

Dividing by $f_{4}rl^{2}$ undoes the multiplication by $f_{4}rl^{2}$.

$$a=\frac{4}{f_{4}rl^{2}}$$

Solve for f_4

$f_{4}=\frac{4}{arl^{2}}$
$l\neq 0\text{ and }r\neq 0\text{ and }a\neq 0$

Steps for Solving Linear Equation

Multiply $l$ and $l$ to get $l^{2}$.

$$\frac{2.5\times 2.5+2\times 3.75+1.5\times 1.5}{2.5+1.5}=f_{4}arl^{2}$$

Multiply $2.5$ and $2.5$ to get $6.25$.

$$\frac{6.25+2\times 3.75+1.5\times 1.5}{2.5+1.5}=f_{4}arl^{2}$$

Multiply $2$ and $3.75$ to get $7.5$.

$$\frac{6.25+7.5+1.5\times 1.5}{2.5+1.5}=f_{4}arl^{2}$$

Add $6.25$ and $7.5$ to get $13.75$.

$$\frac{13.75+1.5\times 1.5}{2.5+1.5}=f_{4}arl^{2}$$

Multiply $1.5$ and $1.5$ to get $2.25$.

$$\frac{13.75+2.25}{2.5+1.5}=f_{4}arl^{2}$$

Add $13.75$ and $2.25$ to get $16$.

$$\frac{16}{2.5+1.5}=f_{4}arl^{2}$$

Add $2.5$ and $1.5$ to get $4$.

$$\frac{16}{4}=f_{4}arl^{2}$$

Divide $16$ by $4$ to get $4$.

$$4=f_{4}arl^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$f_{4}arl^{2}=4$$

The equation is in standard form.

$$arl^{2}f_{4}=4$$

Divide both sides by $arl^{2}$.

$$\frac{arl^{2}f_{4}}{arl^{2}}=\frac{4}{arl^{2}}$$

Dividing by $arl^{2}$ undoes the multiplication by $arl^{2}$.

$$f_{4}=\frac{4}{arl^{2}}$$