Solve (2x-3)/(x-3)=(x-2)/(x-1) | 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

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

Solve for x

$x=\sqrt{3}\approx 1.732050808$
$x=-\sqrt{3}\approx -1.732050808$

Steps by Finding Square Root

Steps Using the Quadratic Formula

Steps by Finding Square Root

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

$$\left(x-1\right)\left(2x-3\right)=\left(x-3\right)\left(x-2\right)$$

Use the distributive property to multiply $x-1$ by $2x-3$ and combine like terms.

$$2x^{2}-5x+3=\left(x-3\right)\left(x-2\right)$$

Use the distributive property to multiply $x-3$ by $x-2$ and combine like terms.

$$2x^{2}-5x+3=x^{2}-5x+6$$

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

$$2x^{2}-5x+3-x^{2}=-5x+6$$

Combine $2x^{2}$ and $-x^{2}$ to get $x^{2}$.

$$x^{2}-5x+3=-5x+6$$

Add $5x$ to both sides.

$$x^{2}-5x+3+5x=6$$

Combine $-5x$ and $5x$ to get $0$.

$$x^{2}+3=6$$

Subtract $3$ from both sides.

$$x^{2}=6-3$$

Subtract $3$ from $6$ to get $3$.

$$x^{2}=3$$

Take the square root of both sides of the equation.

$$x=\sqrt{3}$$ $$x=-\sqrt{3}$$

Steps Using the Quadratic Formula

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

$$\left(x-1\right)\left(2x-3\right)=\left(x-3\right)\left(x-2\right)$$

Use the distributive property to multiply $x-1$ by $2x-3$ and combine like terms.

$$2x^{2}-5x+3=\left(x-3\right)\left(x-2\right)$$

Use the distributive property to multiply $x-3$ by $x-2$ and combine like terms.

$$2x^{2}-5x+3=x^{2}-5x+6$$

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

$$2x^{2}-5x+3-x^{2}=-5x+6$$

Combine $2x^{2}$ and $-x^{2}$ to get $x^{2}$.

$$x^{2}-5x+3=-5x+6$$

Add $5x$ to both sides.

$$x^{2}-5x+3+5x=6$$

Combine $-5x$ and $5x$ to get $0$.

$$x^{2}+3=6$$

Subtract $6$ from both sides.

$$x^{2}+3-6=0$$

Subtract $6$ from $3$ to get $-3$.

$$x^{2}-3=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $0$ for $b$, and $-3$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{0±\sqrt{0^{2}-4\left(-3\right)}}{2}$$

Square $0$.

$$x=\frac{0±\sqrt{-4\left(-3\right)}}{2}$$

Multiply $-4$ times $-3$.

$$x=\frac{0±\sqrt{12}}{2}$$

Take the square root of $12$.

$$x=\frac{0±2\sqrt{3}}{2}$$

Now solve the equation $x=\frac{0±2\sqrt{3}}{2}$ when $±$ is plus.

$$x=\sqrt{3}$$

Now solve the equation $x=\frac{0±2\sqrt{3}}{2}$ when $±$ is minus.

$$x=-\sqrt{3}$$

The equation is now solved.

$$x=\sqrt{3}$$ $$x=-\sqrt{3}$$