Solve for x ax+3x=bx+5

$a x + 3 x = b x + 5$

Subtract $b x$ from both sides of the equation.

$a x + 3 x - b x = 5$

Factor $x$ out of $a x + 3 x - b x$ .

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Factor $x$ out of $a x$ .

$x a + 3 x - b x = 5$

Factor $x$ out of $3 x$ .

$x a + x \cdot 3 - b x = 5$

Factor $x$ out of $- b x$ .

$x a + x \cdot 3 + x (- b) = 5$

Factor $x$ out of $x a + x \cdot 3$ .

$x (a + 3) + x (- b) = 5$

Factor $x$ out of $x (a + 3) + x (- b)$ .

$x (a + 3 - b) = 5$

Divide each term in $x (a + 3 - b) = 5$ by $a + 3 - b$ and simplify.

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Divide each term in $x (a + 3 - b) = 5$ by $a + 3 - b$ .

$\frac{x (a + 3 - b)}{a + 3 - b} = \frac{5}{a + 3 - b}$

Simplify the left side.

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Cancel the common factor of $a + 3 - b$ .

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Cancel the common factor.

$\frac{x (a + 3 - b)}{a + 3 - b} = \frac{5}{a + 3 - b}$

Divide $x$ by $1$ .

$x = \frac{5}{a + 3 - b}$

Solve for x ax+3x=bx+5

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