Solve the System of Equations 6a-4b=3 9a-6b=-1

$6 a - 4 b = 3$ $9 a - 6 b = - 1$

Solve for $a$ in $6 a - 4 b = 3$ .

Add $4 b$ to both sides of the equation.

$6 a = 3 + 4 b$

$9 a - 6 b = - 1$

Divide each term in $6 a = 3 + 4 b$ by $6$ and simplify.

Divide each term in $6 a = 3 + 4 b$ by $6$ .

$\frac{6 a}{6} = \frac{3}{6} + \frac{4 b}{6}$

$9 a - 6 b = - 1$

Simplify the left side.

Cancel the common factor of $6$ .

Cancel the common factor.

$\frac{6 a}{6} = \frac{3}{6} + \frac{4 b}{6}$

$9 a - 6 b = - 1$

Divide $a$ by $1$ .

$a = \frac{3}{6} + \frac{4 b}{6}$

$9 a - 6 b = - 1$

$a = \frac{3}{6} + \frac{4 b}{6}$

$9 a - 6 b = - 1$

$a = \frac{3}{6} + \frac{4 b}{6}$

$9 a - 6 b = - 1$

Simplify the right side.

Simplify each term.

Cancel the common factor of $3$ and $6$ .

Factor $3$ out of $3$ .

$a = \frac{3 (1)}{6} + \frac{4 b}{6}$

$9 a - 6 b = - 1$

Cancel the common factors.

Factor $3$ out of $6$ .

$a = \frac{3 \cdot 1}{3 \cdot 2} + \frac{4 b}{6}$

$9 a - 6 b = - 1$

Cancel the common factor.

$a = \frac{3 \cdot 1}{3 \cdot 2} + \frac{4 b}{6}$

$9 a - 6 b = - 1$

Rewrite the expression.

$a = \frac{1}{2} + \frac{4 b}{6}$

$9 a - 6 b = - 1$

$a = \frac{1}{2} + \frac{4 b}{6}$

$9 a - 6 b = - 1$

$a = \frac{1}{2} + \frac{4 b}{6}$

$9 a - 6 b = - 1$

Cancel the common factor of $4$ and $6$ .

Factor $2$ out of $4 b$ .

$a = \frac{1}{2} + \frac{2 (2 b)}{6}$

$9 a - 6 b = - 1$

Cancel the common factors.

Factor $2$ out of $6$ .

$a = \frac{1}{2} + \frac{2 (2 b)}{2 (3)}$

$9 a - 6 b = - 1$

Cancel the common factor.

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

$9 a - 6 b = - 1$

Rewrite the expression.

$a = \frac{1}{2} + \frac{2 b}{3}$

$9 a - 6 b = - 1$

$a = \frac{1}{2} + \frac{2 b}{3}$

$9 a - 6 b = - 1$

$a = \frac{1}{2} + \frac{2 b}{3}$

$9 a - 6 b = - 1$

$a = \frac{1}{2} + \frac{2 b}{3}$

$9 a - 6 b = - 1$

$a = \frac{1}{2} + \frac{2 b}{3}$

$9 a - 6 b = - 1$

$a = \frac{1}{2} + \frac{2 b}{3}$

$9 a - 6 b = - 1$

$a = \frac{1}{2} + \frac{2 b}{3}$

$9 a - 6 b = - 1$

Replace all occurrences of $a$ with $\frac{1}{2} + \frac{2 b}{3}$ in each equation.

Replace all occurrences of $a$ in $9 a - 6 b = - 1$ with $\frac{1}{2} + \frac{2 b}{3}$ .

$9 (\frac{1}{2} + \frac{2 b}{3}) - 6 b = - 1$

$a = \frac{1}{2} + \frac{2 b}{3}$

Simplify the left side.

Simplify $9 (\frac{1}{2} + \frac{2 b}{3}) - 6 b$ .

Simplify each term.

Apply the distributive property.

$9 (\frac{1}{2}) + 9 (\frac{2 b}{3}) - 6 b = - 1$

$a = \frac{1}{2} + \frac{2 b}{3}$

Combine $9$ and $\frac{1}{2}$ .

$\frac{9}{2} + 9 (\frac{2 b}{3}) - 6 b = - 1$

$a = \frac{1}{2} + \frac{2 b}{3}$

Cancel the common factor of $3$ .

Factor $3$ out of $9$ .

$\frac{9}{2} + 3 (3) (\frac{2 b}{3}) - 6 b = - 1$

$a = \frac{1}{2} + \frac{2 b}{3}$

Cancel the common factor.

$\frac{9}{2} + 3 \cdot (3 (\frac{2 b}{3})) - 6 b = - 1$

$a = \frac{1}{2} + \frac{2 b}{3}$

Rewrite the expression.

$\frac{9}{2} + 3 (2 b) - 6 b = - 1$

$a = \frac{1}{2} + \frac{2 b}{3}$

$\frac{9}{2} + 3 (2 b) - 6 b = - 1$

$a = \frac{1}{2} + \frac{2 b}{3}$

Multiply $2$ by $3$ .

$\frac{9}{2} + 6 b - 6 b = - 1$

$a = \frac{1}{2} + \frac{2 b}{3}$

$\frac{9}{2} + 6 b - 6 b = - 1$

$a = \frac{1}{2} + \frac{2 b}{3}$

Combine the opposite terms in $\frac{9}{2} + 6 b - 6 b$ .

Subtract $6 b$ from $6 b$ .

$\frac{9}{2} + 0 = - 1$

$a = \frac{1}{2} + \frac{2 b}{3}$

Add $\frac{9}{2}$ and $0$ .

$\frac{9}{2} = - 1$

$a = \frac{1}{2} + \frac{2 b}{3}$

$\frac{9}{2} = - 1$

$a = \frac{1}{2} + \frac{2 b}{3}$

$\frac{9}{2} = - 1$

$a = \frac{1}{2} + \frac{2 b}{3}$

$\frac{9}{2} = - 1$

$a = \frac{1}{2} + \frac{2 b}{3}$

$\frac{9}{2} = - 1$

$a = \frac{1}{2} + \frac{2 b}{3}$

Since $\frac{9}{2} = - 1$ is not true, there is no solution.

No solution

Related Questions