Solve {l}{12x=-5y}{24x+7y=-4}

$$\left\{ \begin{array} { l } { 12 x = - 5 y } \\ { 24 x + 7 y = - 4 } \end{array} \right.$$

Solve for x, y

$$x=-\frac{5}{9}\approx -0.555555556$$

$$y = \frac{4}{3} = 1\frac{1}{3} \approx 1.333333333$$

Steps Using Substitution

Consider the first equation. Add 5y to both sides.

$$12x+5y=0$$

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.

$$12x+5y=0,24x+7y=-4$$

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

$$12x+5y=0$$

Subtract 5y from both sides of the equation.

$$12x=-5y$$

Divide both sides by 12.

$$x=\frac{1}{12}\left(-5\right)y$$

Multiply \frac{1}{12} times -5y.

$$x=-\frac{5}{12}y$$

Substitute -\frac{5y}{12} for x in the other equation, 24x+7y=-4.

$$24\left(-\frac{5}{12}\right)y+7y=-4$$

Multiply 24 times -\frac{5y}{12}.

$$-10y+7y=-4$$

Add -10y to 7y.

$$-3y=-4$$

Divide both sides by -3.

$$y=\frac{4}{3}$$

Substitute \frac{4}{3} for y in x=-\frac{5}{12}y. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{5}{12}\times \frac{4}{3}$$

Multiply -\frac{5}{12} times \frac{4}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{5}{9}$$

The system is now solved.

$$x=-\frac{5}{9},y=\frac{4}{3}$$

Steps Using Matrices

Consider the first equation. Add 5y to both sides.

$$12x+5y=0$$

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

$$12x+5y=0,24x+7y=-4$$

Write the equations in matrix form.

$$\left(\begin{matrix}12&5\\24&7\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\-4\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}12&5\\24&7\end{matrix}\right).

$$inverse(\left(\begin{matrix}12&5\\24&7\end{matrix}\right))\left(\begin{matrix}12&5\\24&7\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}12&5\\24&7\end{matrix}\right))\left(\begin{matrix}0\\-4\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}12&5\\24&7\end{matrix}\right))\left(\begin{matrix}0\\-4\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}12&5\\24&7\end{matrix}\right))\left(\begin{matrix}0\\-4\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{7}{12\times 7-5\times 24}&-\frac{5}{12\times 7-5\times 24}\\-\frac{24}{12\times 7-5\times 24}&\frac{12}{12\times 7-5\times 24}\end{matrix}\right)\left(\begin{matrix}0\\-4\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{7}{36}&\frac{5}{36}\\\frac{2}{3}&-\frac{1}{3}\end{matrix}\right)\left(\begin{matrix}0\\-4\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{36}\left(-4\right)\\-\frac{1}{3}\left(-4\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{5}{9}\\\frac{4}{3}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=-\frac{5}{9},y=\frac{4}{3}$$

Steps Using Elimination

Consider the first equation. Add 5y to both sides.

$$12x+5y=0$$

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.

$$12x+5y=0,24x+7y=-4$$

To make 12x and 24x equal, multiply all terms on each side of the first equation by 24 and all terms on each side of the second by 12.

$$24\times 12x+24\times 5y=0,12\times 24x+12\times 7y=12\left(-4\right)$$

Simplify.

$$288x+120y=0,288x+84y=-48$$

Subtract 288x+84y=-48 from 288x+120y=0 by subtracting like terms on each side of the equal sign.

$$288x-288x+120y-84y=48$$

Add 288x to -288x. Terms 288x and -288x cancel out, leaving an equation with only one variable that can be solved.

$$120y-84y=48$$

Add 120y to -84y.

$$36y=48$$

Divide both sides by 36.

$$y=\frac{4}{3}$$

Substitute \frac{4}{3} for y in 24x+7y=-4. Because the resulting equation contains only one variable, you can solve for x directly.

$$24x+7\times \frac{4}{3}=-4$$

Multiply 7 times \frac{4}{3}.

$$24x+\frac{28}{3}=-4$$

Subtract \frac{28}{3} from both sides of the equation.

$$24x=-\frac{40}{3}$$

Divide both sides by 24.

$$x=-\frac{5}{9}$$

The system is now solved.

$$x=-\frac{5}{9},y=\frac{4}{3}$$