Solve (2 - 7x)(4 + 5x) = 1 1-52 3+ 72 ) | 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

$$( \frac { 2 - 7 x } { 1 - 5 x } ) ( \frac { 4 + 5 x } { 3 + 7 x } ) = 1$$

Solve for x

$x=\frac{1}{2}=0.5$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Variable $x$ cannot be equal to any of the values $-\frac{3}{7},\frac{1}{5}$ since division by zero is not defined. Multiply both sides of the equation by $\left(5x-1\right)\left(7x+3\right)$, the least common multiple of $1-5x,3+7x$.

$$\left(-3-7x\right)\left(2-7x\right)\times \frac{4+5x}{3+7x}=\left(5x-1\right)\left(7x+3\right)$$

Express $\left(-3-7x\right)\times \frac{4+5x}{3+7x}$ as a single fraction.

$$\frac{\left(-3-7x\right)\left(4+5x\right)}{3+7x}\left(2-7x\right)=\left(5x-1\right)\left(7x+3\right)$$

Use the distributive property to multiply $\frac{\left(-3-7x\right)\left(4+5x\right)}{3+7x}$ by $2-7x$.

$$2\times \frac{\left(-3-7x\right)\left(4+5x\right)}{3+7x}-7\times \frac{\left(-3-7x\right)\left(4+5x\right)}{3+7x}x=\left(5x-1\right)\left(7x+3\right)$$

Use the distributive property to multiply $-3-7x$ by $4+5x$ and combine like terms.

$$2\times \frac{-12-43x-35x^{2}}{3+7x}-7\times \frac{\left(-3-7x\right)\left(4+5x\right)}{3+7x}x=\left(5x-1\right)\left(7x+3\right)$$

Express $2\times \frac{-12-43x-35x^{2}}{3+7x}$ as a single fraction.

$$\frac{2\left(-12-43x-35x^{2}\right)}{3+7x}-7\times \frac{\left(-3-7x\right)\left(4+5x\right)}{3+7x}x=\left(5x-1\right)\left(7x+3\right)$$

Use the distributive property to multiply $-3-7x$ by $4+5x$ and combine like terms.

$$\frac{2\left(-12-43x-35x^{2}\right)}{3+7x}-7\times \frac{-12-43x-35x^{2}}{3+7x}x=\left(5x-1\right)\left(7x+3\right)$$

Express $-7\times \frac{-12-43x-35x^{2}}{3+7x}$ as a single fraction.

$$\frac{2\left(-12-43x-35x^{2}\right)}{3+7x}+\frac{-7\left(-12-43x-35x^{2}\right)}{3+7x}x=\left(5x-1\right)\left(7x+3\right)$$

Express $\frac{-7\left(-12-43x-35x^{2}\right)}{3+7x}x$ as a single fraction.

$$\frac{2\left(-12-43x-35x^{2}\right)}{3+7x}+\frac{-7\left(-12-43x-35x^{2}\right)x}{3+7x}=\left(5x-1\right)\left(7x+3\right)$$

Since $\frac{2\left(-12-43x-35x^{2}\right)}{3+7x}$ and $\frac{-7\left(-12-43x-35x^{2}\right)x}{3+7x}$ have the same denominator, add them by adding their numerators.

$$\frac{2\left(-12-43x-35x^{2}\right)-7\left(-12-43x-35x^{2}\right)x}{3+7x}=\left(5x-1\right)\left(7x+3\right)$$

Do the multiplications in $2\left(-12-43x-35x^{2}\right)-7\left(-12-43x-35x^{2}\right)x$.

$$\frac{-24-86x-70x^{2}+84x+301x^{2}+245x^{3}}{3+7x}=\left(5x-1\right)\left(7x+3\right)$$

Combine like terms in $-24-86x-70x^{2}+84x+301x^{2}+245x^{3}$.

$$\frac{-24-2x+231x^{2}+245x^{3}}{3+7x}=\left(5x-1\right)\left(7x+3\right)$$

Use the distributive property to multiply $5x-1$ by $7x+3$ and combine like terms.

$$\frac{-24-2x+231x^{2}+245x^{3}}{3+7x}=35x^{2}+8x-3$$

Subtract $35x^{2}$ from both sides.

$$\frac{-24-2x+231x^{2}+245x^{3}}{3+7x}-35x^{2}=8x-3$$

To add or subtract expressions, expand them to make their denominators the same. Multiply $-35x^{2}$ times $\frac{3+7x}{3+7x}$.

$$\frac{-24-2x+231x^{2}+245x^{3}}{3+7x}+\frac{-35x^{2}\left(3+7x\right)}{3+7x}=8x-3$$

Since $\frac{-24-2x+231x^{2}+245x^{3}}{3+7x}$ and $\frac{-35x^{2}\left(3+7x\right)}{3+7x}$ have the same denominator, add them by adding their numerators.

$$\frac{-24-2x+231x^{2}+245x^{3}-35x^{2}\left(3+7x\right)}{3+7x}=8x-3$$

Do the multiplications in $-24-2x+231x^{2}+245x^{3}-35x^{2}\left(3+7x\right)$.

$$\frac{-24-2x+231x^{2}+245x^{3}-105x^{2}-245x^{3}}{3+7x}=8x-3$$

Combine like terms in $-24-2x+231x^{2}+245x^{3}-105x^{2}-245x^{3}$.

$$\frac{-24-2x+126x^{2}}{3+7x}=8x-3$$

Subtract $8x$ from both sides.

$$\frac{-24-2x+126x^{2}}{3+7x}-8x=-3$$

To add or subtract expressions, expand them to make their denominators the same. Multiply $-8x$ times $\frac{3+7x}{3+7x}$.

$$\frac{-24-2x+126x^{2}}{3+7x}+\frac{-8x\left(3+7x\right)}{3+7x}=-3$$

Since $\frac{-24-2x+126x^{2}}{3+7x}$ and $\frac{-8x\left(3+7x\right)}{3+7x}$ have the same denominator, add them by adding their numerators.

$$\frac{-24-2x+126x^{2}-8x\left(3+7x\right)}{3+7x}=-3$$

Do the multiplications in $-24-2x+126x^{2}-8x\left(3+7x\right)$.

$$\frac{-24-2x+126x^{2}-24x-56x^{2}}{3+7x}=-3$$

Combine like terms in $-24-2x+126x^{2}-24x-56x^{2}$.

$$\frac{-24-26x+70x^{2}}{3+7x}=-3$$

Add $3$ to both sides.

$$\frac{-24-26x+70x^{2}}{3+7x}+3=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply $3$ times $\frac{3+7x}{3+7x}$.

$$\frac{-24-26x+70x^{2}}{3+7x}+\frac{3\left(3+7x\right)}{3+7x}=0$$

Since $\frac{-24-26x+70x^{2}}{3+7x}$ and $\frac{3\left(3+7x\right)}{3+7x}$ have the same denominator, add them by adding their numerators.

$$\frac{-24-26x+70x^{2}+3\left(3+7x\right)}{3+7x}=0$$

Do the multiplications in $-24-26x+70x^{2}+3\left(3+7x\right)$.

$$\frac{-24-26x+70x^{2}+9+21x}{3+7x}=0$$

Combine like terms in $-24-26x+70x^{2}+9+21x$.

$$\frac{-15-5x+70x^{2}}{3+7x}=0$$

Variable $x$ cannot be equal to $-\frac{3}{7}$ since division by zero is not defined. Multiply both sides of the equation by $7x+3$.

$$-15-5x+70x^{2}=0$$

All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.

$$70x^{2}-5x-15=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $70$ for $a$, $-5$ for $b$, and $-15$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 70\left(-15\right)}}{2\times 70}$$

Square $-5$.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 70\left(-15\right)}}{2\times 70}$$

Multiply $-4$ times $70$.

$$x=\frac{-\left(-5\right)±\sqrt{25-280\left(-15\right)}}{2\times 70}$$

Multiply $-280$ times $-15$.

$$x=\frac{-\left(-5\right)±\sqrt{25+4200}}{2\times 70}$$

Add $25$ to $4200$.

$$x=\frac{-\left(-5\right)±\sqrt{4225}}{2\times 70}$$

Take the square root of $4225$.

$$x=\frac{-\left(-5\right)±65}{2\times 70}$$

The opposite of $-5$ is $5$.

$$x=\frac{5±65}{2\times 70}$$

Multiply $2$ times $70$.

$$x=\frac{5±65}{140}$$

Now solve the equation $x=\frac{5±65}{140}$ when $±$ is plus. Add $5$ to $65$.

$$x=\frac{70}{140}$$

Reduce the fraction $\frac{70}{140}$ to lowest terms by extracting and canceling out $70$.

$$x=\frac{1}{2}$$

Now solve the equation $x=\frac{5±65}{140}$ when $±$ is minus. Subtract $65$ from $5$.

$$x=-\frac{60}{140}$$

Reduce the fraction $\frac{-60}{140}$ to lowest terms by extracting and canceling out $20$.

$$x=-\frac{3}{7}$$

The equation is now solved.

$$x=\frac{1}{2}$$ $$x=-\frac{3}{7}$$

Variable $x$ cannot be equal to $-\frac{3}{7}$.

$$x=\frac{1}{2}$$

Steps for Completing the Square

Variable $x$ cannot be equal to any of the values $-\frac{3}{7},\frac{1}{5}$ since division by zero is not defined. Multiply both sides of the equation by $\left(5x-1\right)\left(7x+3\right)$, the least common multiple of $1-5x,3+7x$.

$$\left(-3-7x\right)\left(2-7x\right)\times \frac{4+5x}{3+7x}=\left(5x-1\right)\left(7x+3\right)$$

Express $\left(-3-7x\right)\times \frac{4+5x}{3+7x}$ as a single fraction.

$$\frac{\left(-3-7x\right)\left(4+5x\right)}{3+7x}\left(2-7x\right)=\left(5x-1\right)\left(7x+3\right)$$

Use the distributive property to multiply $\frac{\left(-3-7x\right)\left(4+5x\right)}{3+7x}$ by $2-7x$.

$$2\times \frac{\left(-3-7x\right)\left(4+5x\right)}{3+7x}-7\times \frac{\left(-3-7x\right)\left(4+5x\right)}{3+7x}x=\left(5x-1\right)\left(7x+3\right)$$

Use the distributive property to multiply $-3-7x$ by $4+5x$ and combine like terms.

$$2\times \frac{-12-43x-35x^{2}}{3+7x}-7\times \frac{\left(-3-7x\right)\left(4+5x\right)}{3+7x}x=\left(5x-1\right)\left(7x+3\right)$$

Express $2\times \frac{-12-43x-35x^{2}}{3+7x}$ as a single fraction.

$$\frac{2\left(-12-43x-35x^{2}\right)}{3+7x}-7\times \frac{\left(-3-7x\right)\left(4+5x\right)}{3+7x}x=\left(5x-1\right)\left(7x+3\right)$$

Use the distributive property to multiply $-3-7x$ by $4+5x$ and combine like terms.

$$\frac{2\left(-12-43x-35x^{2}\right)}{3+7x}-7\times \frac{-12-43x-35x^{2}}{3+7x}x=\left(5x-1\right)\left(7x+3\right)$$

Express $-7\times \frac{-12-43x-35x^{2}}{3+7x}$ as a single fraction.

$$\frac{2\left(-12-43x-35x^{2}\right)}{3+7x}+\frac{-7\left(-12-43x-35x^{2}\right)}{3+7x}x=\left(5x-1\right)\left(7x+3\right)$$

Express $\frac{-7\left(-12-43x-35x^{2}\right)}{3+7x}x$ as a single fraction.

$$\frac{2\left(-12-43x-35x^{2}\right)}{3+7x}+\frac{-7\left(-12-43x-35x^{2}\right)x}{3+7x}=\left(5x-1\right)\left(7x+3\right)$$

Since $\frac{2\left(-12-43x-35x^{2}\right)}{3+7x}$ and $\frac{-7\left(-12-43x-35x^{2}\right)x}{3+7x}$ have the same denominator, add them by adding their numerators.

$$\frac{2\left(-12-43x-35x^{2}\right)-7\left(-12-43x-35x^{2}\right)x}{3+7x}=\left(5x-1\right)\left(7x+3\right)$$

Do the multiplications in $2\left(-12-43x-35x^{2}\right)-7\left(-12-43x-35x^{2}\right)x$.

$$\frac{-24-86x-70x^{2}+84x+301x^{2}+245x^{3}}{3+7x}=\left(5x-1\right)\left(7x+3\right)$$

Combine like terms in $-24-86x-70x^{2}+84x+301x^{2}+245x^{3}$.

$$\frac{-24-2x+231x^{2}+245x^{3}}{3+7x}=\left(5x-1\right)\left(7x+3\right)$$

Use the distributive property to multiply $5x-1$ by $7x+3$ and combine like terms.

$$\frac{-24-2x+231x^{2}+245x^{3}}{3+7x}=35x^{2}+8x-3$$

Subtract $35x^{2}$ from both sides.

$$\frac{-24-2x+231x^{2}+245x^{3}}{3+7x}-35x^{2}=8x-3$$

To add or subtract expressions, expand them to make their denominators the same. Multiply $-35x^{2}$ times $\frac{3+7x}{3+7x}$.

$$\frac{-24-2x+231x^{2}+245x^{3}}{3+7x}+\frac{-35x^{2}\left(3+7x\right)}{3+7x}=8x-3$$

Since $\frac{-24-2x+231x^{2}+245x^{3}}{3+7x}$ and $\frac{-35x^{2}\left(3+7x\right)}{3+7x}$ have the same denominator, add them by adding their numerators.

$$\frac{-24-2x+231x^{2}+245x^{3}-35x^{2}\left(3+7x\right)}{3+7x}=8x-3$$

Do the multiplications in $-24-2x+231x^{2}+245x^{3}-35x^{2}\left(3+7x\right)$.

$$\frac{-24-2x+231x^{2}+245x^{3}-105x^{2}-245x^{3}}{3+7x}=8x-3$$

Combine like terms in $-24-2x+231x^{2}+245x^{3}-105x^{2}-245x^{3}$.

$$\frac{-24-2x+126x^{2}}{3+7x}=8x-3$$

Subtract $8x$ from both sides.

$$\frac{-24-2x+126x^{2}}{3+7x}-8x=-3$$

To add or subtract expressions, expand them to make their denominators the same. Multiply $-8x$ times $\frac{3+7x}{3+7x}$.

$$\frac{-24-2x+126x^{2}}{3+7x}+\frac{-8x\left(3+7x\right)}{3+7x}=-3$$

Since $\frac{-24-2x+126x^{2}}{3+7x}$ and $\frac{-8x\left(3+7x\right)}{3+7x}$ have the same denominator, add them by adding their numerators.

$$\frac{-24-2x+126x^{2}-8x\left(3+7x\right)}{3+7x}=-3$$

Do the multiplications in $-24-2x+126x^{2}-8x\left(3+7x\right)$.

$$\frac{-24-2x+126x^{2}-24x-56x^{2}}{3+7x}=-3$$

Combine like terms in $-24-2x+126x^{2}-24x-56x^{2}$.

$$\frac{-24-26x+70x^{2}}{3+7x}=-3$$

Variable $x$ cannot be equal to $-\frac{3}{7}$ since division by zero is not defined. Multiply both sides of the equation by $7x+3$.

$$-24-26x+70x^{2}=-3\left(7x+3\right)$$

Use the distributive property to multiply $-3$ by $7x+3$.

$$-24-26x+70x^{2}=-21x-9$$

Add $21x$ to both sides.

$$-24-26x+70x^{2}+21x=-9$$

Combine $-26x$ and $21x$ to get $-5x$.

$$-24-5x+70x^{2}=-9$$

Add $24$ to both sides.

$$-5x+70x^{2}=-9+24$$

Add $-9$ and $24$ to get $15$.

$$-5x+70x^{2}=15$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form $x^{2}+bx=c$.

$$70x^{2}-5x=15$$

Divide both sides by $70$.

$$\frac{70x^{2}-5x}{70}=\frac{15}{70}$$

Dividing by $70$ undoes the multiplication by $70$.

$$x^{2}+\left(-\frac{5}{70}\right)x=\frac{15}{70}$$

Reduce the fraction $\frac{-5}{70}$ to lowest terms by extracting and canceling out $5$.

$$x^{2}-\frac{1}{14}x=\frac{15}{70}$$

Reduce the fraction $\frac{15}{70}$ to lowest terms by extracting and canceling out $5$.

$$x^{2}-\frac{1}{14}x=\frac{3}{14}$$

Divide $-\frac{1}{14}$, the coefficient of the $x$ term, by $2$ to get $-\frac{1}{28}$. Then add the square of $-\frac{1}{28}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{14}x+\left(-\frac{1}{28}\right)^{2}=\frac{3}{14}+\left(-\frac{1}{28}\right)^{2}$$

Square $-\frac{1}{28}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{14}x+\frac{1}{784}=\frac{3}{14}+\frac{1}{784}$$

Add $\frac{3}{14}$ to $\frac{1}{784}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{14}x+\frac{1}{784}=\frac{169}{784}$$

Factor $x^{2}-\frac{1}{14}x+\frac{1}{784}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{1}{28}\right)^{2}=\frac{169}{784}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{28}\right)^{2}}=\sqrt{\frac{169}{784}}$$

Simplify.

$$x-\frac{1}{28}=\frac{13}{28}$$ $$x-\frac{1}{28}=-\frac{13}{28}$$

Add $\frac{1}{28}$ to both sides of the equation.

$$x=\frac{1}{2}$$ $$x=-\frac{3}{7}$$

Variable $x$ cannot be equal to $-\frac{3}{7}$.

$$x=\frac{1}{2}$$