Solve {l}{6x-4y=30}{12x+5y=-18}

$$\left. \begin{array} { l } { 6 x - 4 y = 30 } \\ { 12 x + 5 y = - 18 } \end{array} \right.$$

Solve for x, y

$$x=1$$

$$y=-6$$

Steps Using Substitution

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

$$6x-4y=30,12x+5y=-18$$

Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.

$$6x-4y=30$$

Add 4y to both sides of the equation.

$$6x=4y+30$$

Divide both sides by 6.

$$x=\frac{1}{6}\left(4y+30\right)$$

Multiply \frac{1}{6} times 4y+30.

$$x=\frac{2}{3}y+5$$

Substitute \frac{2y}{3}+5 for x in the other equation, 12x+5y=-18.

$$12\left(\frac{2}{3}y+5\right)+5y=-18$$

Multiply 12 times \frac{2y}{3}+5.

$$8y+60+5y=-18$$

Add 8y to 5y.

$$13y+60=-18$$

Subtract 60 from both sides of the equation.

$$13y=-78$$

Divide both sides by 13.

$$y=-6$$

Substitute -6 for y in x=\frac{2}{3}y+5. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{2}{3}\left(-6\right)+5$$

Multiply \frac{2}{3} times -6.

$$x=-4+5$$

Add 5 to -4.

$$x=1$$

The system is now solved.

$$x=1,y=-6$$

Steps Using Matrices

Put the equations in standard form and then use matrices to solve the system of equations.

$$6x-4y=30,12x+5y=-18$$

Write the equations in matrix form.

$$\left(\begin{matrix}6&-4\\12&5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}30\\-18\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}6&-4\\12&5\end{matrix}\right).

$$inverse(\left(\begin{matrix}6&-4\\12&5\end{matrix}\right))\left(\begin{matrix}6&-4\\12&5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}6&-4\\12&5\end{matrix}\right))\left(\begin{matrix}30\\-18\end{matrix}\right)$$

The product of a matrix and its inverse is the identity matrix.

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}6&-4\\12&5\end{matrix}\right))\left(\begin{matrix}30\\-18\end{matrix}\right)$$

Multiply the matrices on the left hand side of the equal sign.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}6&-4\\12&5\end{matrix}\right))\left(\begin{matrix}30\\-18\end{matrix}\right)$$

For the 2\times 2 matrix \left(\begin{matrix}a&b\\c&d\end{matrix}\right), the inverse matrix is \left(\begin{matrix}\frac{d}{ad-bc}&\frac{-b}{ad-bc}\\\frac{-c}{ad-bc}&\frac{a}{ad-bc}\end{matrix}\right), so the matrix equation can be rewritten as a matrix multiplication problem.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{6\times 5-\left(-4\times 12\right)}&-\frac{-4}{6\times 5-\left(-4\times 12\right)}\\-\frac{12}{6\times 5-\left(-4\times 12\right)}&\frac{6}{6\times 5-\left(-4\times 12\right)}\end{matrix}\right)\left(\begin{matrix}30\\-18\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{78}&\frac{2}{39}\\-\frac{2}{13}&\frac{1}{13}\end{matrix}\right)\left(\begin{matrix}30\\-18\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{78}\times 30+\frac{2}{39}\left(-18\right)\\-\frac{2}{13}\times 30+\frac{1}{13}\left(-18\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}1\\-6\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=1,y=-6$$

Steps Using Elimination

In order to solve by elimination, coefficients of one of the variables must be the same in both equations so that the variable will cancel out when one equation is subtracted from the other.

$$6x-4y=30,12x+5y=-18$$

To make 6x and 12x equal, multiply all terms on each side of the first equation by 12 and all terms on each side of the second by 6.

$$12\times 6x+12\left(-4\right)y=12\times 30,6\times 12x+6\times 5y=6\left(-18\right)$$

Simplify.

$$72x-48y=360,72x+30y=-108$$

Subtract 72x+30y=-108 from 72x-48y=360 by subtracting like terms on each side of the equal sign.

$$72x-72x-48y-30y=360+108$$

Add 72x to -72x. Terms 72x and -72x cancel out, leaving an equation with only one variable that can be solved.

$$-48y-30y=360+108$$

Add -48y to -30y.

$$-78y=360+108$$

Add 360 to 108.

$$-78y=468$$

Divide both sides by -78.

$$y=-6$$

Substitute -6 for y in 12x+5y=-18. Because the resulting equation contains only one variable, you can solve for x directly.

$$12x+5\left(-6\right)=-18$$

Multiply 5 times -6.

$$12x-30=-18$$

Add 30 to both sides of the equation.

$$12x=12$$

Divide both sides by 12.

$$x=1$$

The system is now solved.

$$x=1,y=-6$$