Solve (3P - 3)/4 * (2P * 3)/3 = 3/3 * P | 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

$$\frac{3P-3}{4}\cdot\frac{2P*3}{3}=\frac{3}{3}\cdot P$$

Solve for P

$P = \frac{5}{3} = 1\frac{2}{3} \approx 1.666666667$
$P=0$

Steps Using Factoring

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring

Multiply both sides of the equation by $12$, the least common multiple of $4,3$.

$$\left(3P-3\right)\times 2P\times 3=12P$$

Multiply $2$ and $3$ to get $6$.

$$\left(3P-3\right)\times 6P=12P$$

Use the distributive property to multiply $3P-3$ by $6$.

$$\left(18P-18\right)P=12P$$

Use the distributive property to multiply $18P-18$ by $P$.

$$18P^{2}-18P=12P$$

Subtract $12P$ from both sides.

$$18P^{2}-18P-12P=0$$

Combine $-18P$ and $-12P$ to get $-30P$.

$$18P^{2}-30P=0$$

Factor out $P$.

$$P\left(18P-30\right)=0$$

To find equation solutions, solve $P=0$ and $18P-30=0$.

$$P=0$$ $$P=\frac{5}{3}$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by $12$, the least common multiple of $4,3$.

$$\left(3P-3\right)\times 2P\times 3=12P$$

Multiply $2$ and $3$ to get $6$.

$$\left(3P-3\right)\times 6P=12P$$

Use the distributive property to multiply $3P-3$ by $6$.

$$\left(18P-18\right)P=12P$$

Use the distributive property to multiply $18P-18$ by $P$.

$$18P^{2}-18P=12P$$

Subtract $12P$ from both sides.

$$18P^{2}-18P-12P=0$$

Combine $-18P$ and $-12P$ to get $-30P$.

$$18P^{2}-30P=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $18$ for $a$, $-30$ for $b$, and $0$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$P=\frac{-\left(-30\right)±\sqrt{\left(-30\right)^{2}}}{2\times 18}$$

Take the square root of $\left(-30\right)^{2}$.

$$P=\frac{-\left(-30\right)±30}{2\times 18}$$

The opposite of $-30$ is $30$.

$$P=\frac{30±30}{2\times 18}$$

Multiply $2$ times $18$.

$$P=\frac{30±30}{36}$$

Now solve the equation $P=\frac{30±30}{36}$ when $±$ is plus. Add $30$ to $30$.

$$P=\frac{60}{36}$$

Reduce the fraction $\frac{60}{36}$ to lowest terms by extracting and canceling out $12$.

$$P=\frac{5}{3}$$

Now solve the equation $P=\frac{30±30}{36}$ when $±$ is minus. Subtract $30$ from $30$.

$$P=\frac{0}{36}$$

Divide $0$ by $36$.

$$P=0$$

The equation is now solved.

$$P=\frac{5}{3}$$ $$P=0$$

Steps for Completing the Square

Multiply both sides of the equation by $12$, the least common multiple of $4,3$.

$$\left(3P-3\right)\times 2P\times 3=12P$$

Multiply $2$ and $3$ to get $6$.

$$\left(3P-3\right)\times 6P=12P$$

Use the distributive property to multiply $3P-3$ by $6$.

$$\left(18P-18\right)P=12P$$

Use the distributive property to multiply $18P-18$ by $P$.

$$18P^{2}-18P=12P$$

Subtract $12P$ from both sides.

$$18P^{2}-18P-12P=0$$

Combine $-18P$ and $-12P$ to get $-30P$.

$$18P^{2}-30P=0$$

Divide both sides by $18$.

$$\frac{18P^{2}-30P}{18}=\frac{0}{18}$$

Dividing by $18$ undoes the multiplication by $18$.

$$P^{2}+\left(-\frac{30}{18}\right)P=\frac{0}{18}$$

Reduce the fraction $\frac{-30}{18}$ to lowest terms by extracting and canceling out $6$.

$$P^{2}-\frac{5}{3}P=\frac{0}{18}$$

Divide $0$ by $18$.

$$P^{2}-\frac{5}{3}P=0$$

Divide $-\frac{5}{3}$, the coefficient of the $x$ term, by $2$ to get $-\frac{5}{6}$. Then add the square of $-\frac{5}{6}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$P^{2}-\frac{5}{3}P+\left(-\frac{5}{6}\right)^{2}=\left(-\frac{5}{6}\right)^{2}$$

Square $-\frac{5}{6}$ by squaring both the numerator and the denominator of the fraction.

$$P^{2}-\frac{5}{3}P+\frac{25}{36}=\frac{25}{36}$$

Factor $P^{2}-\frac{5}{3}P+\frac{25}{36}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(P-\frac{5}{6}\right)^{2}=\frac{25}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(P-\frac{5}{6}\right)^{2}}=\sqrt{\frac{25}{36}}$$

Simplify.

$$P-\frac{5}{6}=\frac{5}{6}$$ $$P-\frac{5}{6}=-\frac{5}{6}$$

Add $\frac{5}{6}$ to both sides of the equation.

$$P=\frac{5}{3}$$ $$P=0$$