Solve in Terms of the Arbitrary Variable x 2x+3y=3 ax+by=6

$2 x + 3 y = 3$ $a x + b y = 6$

Solve the equation for $a$ .

Simplify each term.

Apply the distributive property.

$a x + b \cdot 1 + b (- \frac{2 x}{3}) = 6$

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

Multiply $b$ by $1$ .

$a x + b + b (- \frac{2 x}{3}) = 6$

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

Rewrite using the commutative property of multiplication.

$a x + b - b \frac{2 x}{3} = 6$

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

Combine $\frac{2 x}{3}$ and $b$ .

$a x + b - \frac{2 x b}{3} = 6$

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

$a x + b - \frac{2 x b}{3} = 6$

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

Move all terms not containing $a$ to the right side of the equation.

Subtract $b$ from both sides of the equation.

$a x - \frac{2 x b}{3} = 6 - b$

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

Add $\frac{2 x b}{3}$ to both sides of the equation.

$a x = 6 - b + \frac{2 x b}{3}$

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

$a x = 6 - b + \frac{2 x b}{3}$

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

Divide each term in $a x = 6 - b + \frac{2 x b}{3}$ by $x$ and simplify.

Divide each term in $a x = 6 - b + \frac{2 x b}{3}$ by $x$ .

$\frac{a x}{x} = \frac{6}{x} + \frac{- b}{x} + \frac{\frac{2 x b}{3}}{x}$

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

Simplify the left side.

Cancel the common factor of $x$ .

Cancel the common factor.

$\frac{a x}{x} = \frac{6}{x} + \frac{- b}{x} + \frac{\frac{2 x b}{3}}{x}$

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

Divide $a$ by $1$ .

$a = \frac{6}{x} + \frac{- b}{x} + \frac{\frac{2 x b}{3}}{x}$

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

$a = \frac{6}{x} + \frac{- b}{x} + \frac{\frac{2 x b}{3}}{x}$

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

$a = \frac{6}{x} + \frac{- b}{x} + \frac{\frac{2 x b}{3}}{x}$

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

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

$a = \frac{6}{x} - \frac{b}{x} + \frac{\frac{2 x b}{3}}{x}$

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

Multiply the numerator by the reciprocal of the denominator.

$a = \frac{6}{x} - \frac{b}{x} + \frac{2 x b}{3} \cdot \frac{1}{x}$

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

Cancel the common factor of $x$ .

Factor $x$ out of $2 x b$ .

$a = \frac{6}{x} - \frac{b}{x} + \frac{x (2 b)}{3} \cdot \frac{1}{x}$

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

Cancel the common factor.

$a = \frac{6}{x} - \frac{b}{x} + \frac{x (2 b)}{3} \cdot \frac{1}{x}$

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

Rewrite the expression.

$a = \frac{6}{x} - \frac{b}{x} + \frac{2 b}{3}$

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

$a = \frac{6}{x} - \frac{b}{x} + \frac{2 b}{3}$

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

$a = \frac{6}{x} - \frac{b}{x} + \frac{2 b}{3}$

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

$a = \frac{6}{x} - \frac{b}{x} + \frac{2 b}{3}$

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

$a = \frac{6}{x} - \frac{b}{x} + \frac{2 b}{3}$

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

$a = \frac{6}{x} - \frac{b}{x} + \frac{2 b}{3}$

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

Solve the equation for $b$ .

Simplify $(\frac{6}{x} - \frac{b}{x} + \frac{2 b}{3}) \cdot x + b (1 - \frac{2 x}{3})$ .

Simplify each term.

Apply the distributive property.

$\frac{6}{x} \cdot x - \frac{b}{x} \cdot x + \frac{2 b}{3} \cdot x + b (1 - \frac{2 x}{3}) = 6$

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

Simplify.

Cancel the common factor of $x$ .

Cancel the common factor.

$\frac{6}{x} \cdot x - \frac{b}{x} \cdot x + \frac{2 b}{3} \cdot x + b (1 - \frac{2 x}{3}) = 6$

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

Rewrite the expression.

$6 - \frac{b}{x} \cdot x + \frac{2 b}{3} \cdot x + b (1 - \frac{2 x}{3}) = 6$

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

$6 - \frac{b}{x} \cdot x + \frac{2 b}{3} \cdot x + b (1 - \frac{2 x}{3}) = 6$

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

Cancel the common factor of $x$ .

Move the leading negative in $- \frac{b}{x}$ into the numerator.

$6 + \frac{- b}{x} \cdot x + \frac{2 b}{3} \cdot x + b (1 - \frac{2 x}{3}) = 6$

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

Cancel the common factor.

$6 + \frac{- b}{x} \cdot x + \frac{2 b}{3} \cdot x + b (1 - \frac{2 x}{3}) = 6$

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

Rewrite the expression.

$6 - b + \frac{2 b}{3} \cdot x + b (1 - \frac{2 x}{3}) = 6$

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

$6 - b + \frac{2 b}{3} \cdot x + b (1 - \frac{2 x}{3}) = 6$

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

Combine $\frac{2 b}{3}$ and $x$ .

$6 - b + \frac{2 b x}{3} + b (1 - \frac{2 x}{3}) = 6$

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

$6 - b + \frac{2 b x}{3} + b (1 - \frac{2 x}{3}) = 6$

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

Apply the distributive property.

$6 - b + \frac{2 b x}{3} + b \cdot 1 + b (- \frac{2 x}{3}) = 6$

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

Multiply $b$ by $1$ .

$6 - b + \frac{2 b x}{3} + b + b (- \frac{2 x}{3}) = 6$

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

Rewrite using the commutative property of multiplication.

$6 - b + \frac{2 b x}{3} + b - b \frac{2 x}{3} = 6$

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

Combine $\frac{2 x}{3}$ and $b$ .

$6 - b + \frac{2 b x}{3} + b - \frac{2 x b}{3} = 6$

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

$6 - b + \frac{2 b x}{3} + b - \frac{2 x b}{3} = 6$

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

Simplify terms.

Combine the opposite terms in $6 - b + \frac{2 b x}{3} + b - \frac{2 x b}{3}$ .

Add $- b$ and $b$ .

$6 + \frac{2 b x}{3} + 0 - \frac{2 x b}{3} = 6$

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

Add $6 + \frac{2 b x}{3}$ and $0$ .

$6 + \frac{2 b x}{3} - \frac{2 x b}{3} = 6$

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

$6 + \frac{2 b x}{3} - \frac{2 x b}{3} = 6$

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

Combine the numerators over the common denominator.

$6 + \frac{2 b x - 2 x b}{3} = 6$

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

Combine the opposite terms in $2 b x - 2 x b$ .

Reorder the factors in the terms $2 b x$ and $- 2 x b$ .

$6 + \frac{2 b x - 2 b x}{3} = 6$

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

Subtract $2 b x$ from $2 b x$ .

$6 + \frac{0}{3} = 6$

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

$6 + \frac{0}{3} = 6$

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

Simplify the expression.

Divide $0$ by $3$ .

$6 + 0 = 6$

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

Add $6$ and $0$ .

$6 = 6$

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

$6 = 6$

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

$6 = 6$

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

$6 = 6$

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

Since $6 = 6$ , the equation will always be true.

Always true

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

Always true

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

The simplified system is the arbitrary solution of the original system of equations.

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

Always true

Simplify the right side.

Reorder $1$ and $- \frac{2 x}{3}$ .

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

Always true

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

Always true

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