Solve z_{1} + p_{1}/(pg) + (v_{1} ^ 2)/(2g) = z_{2} + p_{2}/(pg) + (v_{2} ^ 2)/(2g) на | 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

$$z _ { 1 } + \frac { R _ { 1 } } { p \cdot g } + \frac { 1 \frac { 1 } { 1 } } { 2 g } = z _ { 2 } + \frac { p _ { 2 } } { p \cdot g } + \frac { 1 \frac { 2 } { 2 } } { 2 g }$$

Solve for R_1

$R_{1}=p_{2}+gpz_{2}-gpz_{1}$
$p\neq 0\text{ and }g\neq 0$

Steps for Solving Linear Equation

Multiply both sides of the equation by $2gp$, the least common multiple of $pg,2g$.

$$2gpz_{1}+2R_{1}+p\times \frac{1\times 1+1}{1}=2gpz_{2}+2p_{2}+p\times \frac{1\times 2+2}{2}$$

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

$$2gpz_{1}+2R_{1}+p\times \frac{1+1}{1}=2gpz_{2}+2p_{2}+p\times \frac{1\times 2+2}{2}$$

Add $1$ and $1$ to get $2$.

$$2gpz_{1}+2R_{1}+p\times \frac{2}{1}=2gpz_{2}+2p_{2}+p\times \frac{1\times 2+2}{2}$$

Anything divided by one gives itself.

$$2gpz_{1}+2R_{1}+p\times 2=2gpz_{2}+2p_{2}+p\times \frac{1\times 2+2}{2}$$

Multiply $1$ and $2$ to get $2$.

$$2gpz_{1}+2R_{1}+p\times 2=2gpz_{2}+2p_{2}+p\times \frac{2+2}{2}$$

Add $2$ and $2$ to get $4$.

$$2gpz_{1}+2R_{1}+p\times 2=2gpz_{2}+2p_{2}+p\times \frac{4}{2}$$

Divide $4$ by $2$ to get $2$.

$$2gpz_{1}+2R_{1}+p\times 2=2gpz_{2}+2p_{2}+p\times 2$$

Subtract $2gpz_{1}$ from both sides.

$$2R_{1}+p\times 2=2gpz_{2}+2p_{2}+p\times 2-2gpz_{1}$$

Subtract $p\times 2$ from both sides.

$$2R_{1}=2gpz_{2}+2p_{2}+p\times 2-2gpz_{1}-p\times 2$$

Combine $p\times 2$ and $-p\times 2$ to get $0$.

$$2R_{1}=2gpz_{2}+2p_{2}-2gpz_{1}$$

The equation is in standard form.

$$2R_{1}=2p_{2}+2gpz_{2}-2gpz_{1}$$

Divide both sides by $2$.

$$\frac{2R_{1}}{2}=\frac{2p_{2}+2gpz_{2}-2gpz_{1}}{2}$$

Dividing by $2$ undoes the multiplication by $2$.

$$R_{1}=\frac{2p_{2}+2gpz_{2}-2gpz_{1}}{2}$$

Divide $2gpz_{2}+2p_{2}-2gpz_{1}$ by $2$.

$$R_{1}=p_{2}+gpz_{2}-gpz_{1}$$

Solve for g

$\left\{\begin{matrix}g=-\frac{R_{1}-p_{2}}{p\left(z_{1}-z_{2}\right)}\text{, }&R_{1}\neq p_{2}\text{ and }z_{1}\neq z_{2}\text{ and }p\neq 0\\g\neq 0\text{, }&R_{1}=p_{2}\text{ and }z_{1}=z_{2}\text{ and }p\neq 0\end{matrix}\right.$

Steps for Solving Linear Equation

Variable $g$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $2gp$, the least common multiple of $pg,2g$.

$$2gpz_{1}+2R_{1}+p\times \frac{1\times 1+1}{1}=2gpz_{2}+2p_{2}+p\times \frac{1\times 2+2}{2}$$

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

$$2gpz_{1}+2R_{1}+p\times \frac{1+1}{1}=2gpz_{2}+2p_{2}+p\times \frac{1\times 2+2}{2}$$

Add $1$ and $1$ to get $2$.

$$2gpz_{1}+2R_{1}+p\times \frac{2}{1}=2gpz_{2}+2p_{2}+p\times \frac{1\times 2+2}{2}$$

Anything divided by one gives itself.

$$2gpz_{1}+2R_{1}+p\times 2=2gpz_{2}+2p_{2}+p\times \frac{1\times 2+2}{2}$$

Multiply $1$ and $2$ to get $2$.

$$2gpz_{1}+2R_{1}+p\times 2=2gpz_{2}+2p_{2}+p\times \frac{2+2}{2}$$

Add $2$ and $2$ to get $4$.

$$2gpz_{1}+2R_{1}+p\times 2=2gpz_{2}+2p_{2}+p\times \frac{4}{2}$$

Divide $4$ by $2$ to get $2$.

$$2gpz_{1}+2R_{1}+p\times 2=2gpz_{2}+2p_{2}+p\times 2$$

Subtract $2gpz_{2}$ from both sides.

$$2gpz_{1}+2R_{1}+p\times 2-2gpz_{2}=2p_{2}+p\times 2$$

Subtract $2R_{1}$ from both sides.

$$2gpz_{1}+p\times 2-2gpz_{2}=2p_{2}+p\times 2-2R_{1}$$

Subtract $p\times 2$ from both sides.

$$2gpz_{1}-2gpz_{2}=2p_{2}+p\times 2-2R_{1}-p\times 2$$

Combine $p\times 2$ and $-p\times 2$ to get $0$.

$$2gpz_{1}-2gpz_{2}=2p_{2}-2R_{1}$$

Combine all terms containing $g$.

$$\left(2pz_{1}-2pz_{2}\right)g=2p_{2}-2R_{1}$$

Divide both sides by $2pz_{1}-2pz_{2}$.

$$\frac{\left(2pz_{1}-2pz_{2}\right)g}{2pz_{1}-2pz_{2}}=\frac{2p_{2}-2R_{1}}{2pz_{1}-2pz_{2}}$$

Dividing by $2pz_{1}-2pz_{2}$ undoes the multiplication by $2pz_{1}-2pz_{2}$.

$$g=\frac{2p_{2}-2R_{1}}{2pz_{1}-2pz_{2}}$$

Divide $2p_{2}-2R_{1}$ by $2pz_{1}-2pz_{2}$.

$$g=\frac{p_{2}-R_{1}}{p\left(z_{1}-z_{2}\right)}$$

Variable $g$ cannot be equal to $0$.

$$g=\frac{p_{2}-R_{1}}{p\left(z_{1}-z_{2}\right)}\text{, }g\neq 0$$