Solve {l}{-4-y/3=-x/2}{y/6=2/5x+3frac{1}{2}}

$$\left. \begin{array} { l } { - 4 - \frac { y } { 3 } = - \frac { x } { 2 } } \\ { \frac { y } { 6 } = \frac { 2 } { 5 } x + 3 \frac { 1 } { 2 } } \end{array} \right.$$

Solve for y, x

$$x = -\frac{110}{3} = -36\frac{2}{3} \approx -36.666666667$$

$$y=-67$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 6, the least common multiple of 3,2.

$$-24-2y=-3x$$

Add 3x to both sides.

$$-24-2y+3x=0$$

Add 24 to both sides. Anything plus zero gives itself.

$$-2y+3x=24$$

Consider the second equation. Multiply both sides of the equation by 30, the least common multiple of 6,5,2.

$$5y=12x+15\left(3\times 2+1\right)$$

Multiply 3 and 2 to get 6.

$$5y=12x+15\left(6+1\right)$$

Add 6 and 1 to get 7.

$$5y=12x+15\times 7$$

Multiply 15 and 7 to get 105.

$$5y=12x+105$$

Subtract 12x from both sides.

$$5y-12x=105$$

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.

$$-2y+3x=24,5y-12x=105$$

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

$$-2y+3x=24$$

Subtract 3x from both sides of the equation.

$$-2y=-3x+24$$

Divide both sides by -2.

$$y=-\frac{1}{2}\left(-3x+24\right)$$

Multiply -\frac{1}{2} times -3x+24.

$$y=\frac{3}{2}x-12$$

Substitute \frac{3x}{2}-12 for y in the other equation, 5y-12x=105.

$$5\left(\frac{3}{2}x-12\right)-12x=105$$

Multiply 5 times \frac{3x}{2}-12.

$$\frac{15}{2}x-60-12x=105$$

Add \frac{15x}{2} to -12x.

$$-\frac{9}{2}x-60=105$$

Add 60 to both sides of the equation.

$$-\frac{9}{2}x=165$$

Divide both sides of the equation by -\frac{9}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$x=-\frac{110}{3}$$

Substitute -\frac{110}{3} for x in y=\frac{3}{2}x-12. Because the resulting equation contains only one variable, you can solve for y directly.

$$y=\frac{3}{2}\left(-\frac{110}{3}\right)-12$$

Multiply \frac{3}{2} times -\frac{110}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$y=-55-12$$

Add -12 to -55.

$$y=-67$$

The system is now solved.

$$y=-67,x=-\frac{110}{3}$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 6, the least common multiple of 3,2.

$$-24-2y=-3x$$

Add 3x to both sides.

$$-24-2y+3x=0$$

Add 24 to both sides. Anything plus zero gives itself.

$$-2y+3x=24$$

Consider the second equation. Multiply both sides of the equation by 30, the least common multiple of 6,5,2.

$$5y=12x+15\left(3\times 2+1\right)$$

Multiply 3 and 2 to get 6.

$$5y=12x+15\left(6+1\right)$$

Add 6 and 1 to get 7.

$$5y=12x+15\times 7$$

Multiply 15 and 7 to get 105.

$$5y=12x+105$$

Subtract 12x from both sides.

$$5y-12x=105$$

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

$$-2y+3x=24,5y-12x=105$$

Write the equations in matrix form.

$$\left(\begin{matrix}-2&3\\5&-12\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}24\\105\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}-2&3\\5&-12\end{matrix}\right).

$$inverse(\left(\begin{matrix}-2&3\\5&-12\end{matrix}\right))\left(\begin{matrix}-2&3\\5&-12\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}-2&3\\5&-12\end{matrix}\right))\left(\begin{matrix}24\\105\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}-2&3\\5&-12\end{matrix}\right))\left(\begin{matrix}24\\105\end{matrix}\right)$$

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

$$\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}-2&3\\5&-12\end{matrix}\right))\left(\begin{matrix}24\\105\end{matrix}\right)$$

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.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-\frac{12}{-2\left(-12\right)-3\times 5}&-\frac{3}{-2\left(-12\right)-3\times 5}\\-\frac{5}{-2\left(-12\right)-3\times 5}&-\frac{2}{-2\left(-12\right)-3\times 5}\end{matrix}\right)\left(\begin{matrix}24\\105\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-\frac{4}{3}&-\frac{1}{3}\\-\frac{5}{9}&-\frac{2}{9}\end{matrix}\right)\left(\begin{matrix}24\\105\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-\frac{4}{3}\times 24-\frac{1}{3}\times 105\\-\frac{5}{9}\times 24-\frac{2}{9}\times 105\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-67\\-\frac{110}{3}\end{matrix}\right)$$

Extract the matrix elements y and x.

$$y=-67,x=-\frac{110}{3}$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 6, the least common multiple of 3,2.

$$-24-2y=-3x$$

Add 3x to both sides.

$$-24-2y+3x=0$$

Add 24 to both sides. Anything plus zero gives itself.

$$-2y+3x=24$$

Consider the second equation. Multiply both sides of the equation by 30, the least common multiple of 6,5,2.

$$5y=12x+15\left(3\times 2+1\right)$$

Multiply 3 and 2 to get 6.

$$5y=12x+15\left(6+1\right)$$

Add 6 and 1 to get 7.

$$5y=12x+15\times 7$$

Multiply 15 and 7 to get 105.

$$5y=12x+105$$

Subtract 12x from both sides.

$$5y-12x=105$$

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.

$$-2y+3x=24,5y-12x=105$$

To make -2y and 5y equal, multiply all terms on each side of the first equation by 5 and all terms on each side of the second by -2.

$$5\left(-2\right)y+5\times 3x=5\times 24,-2\times 5y-2\left(-12\right)x=-2\times 105$$

Simplify.

$$-10y+15x=120,-10y+24x=-210$$

Subtract -10y+24x=-210 from -10y+15x=120 by subtracting like terms on each side of the equal sign.

$$-10y+10y+15x-24x=120+210$$

Add -10y to 10y. Terms -10y and 10y cancel out, leaving an equation with only one variable that can be solved.

$$15x-24x=120+210$$

Add 15x to -24x.

$$-9x=120+210$$

Add 120 to 210.

$$-9x=330$$

Divide both sides by -9.

$$x=-\frac{110}{3}$$

Substitute -\frac{110}{3} for x in 5y-12x=105. Because the resulting equation contains only one variable, you can solve for y directly.

$$5y-12\left(-\frac{110}{3}\right)=105$$

Multiply -12 times -\frac{110}{3}.

$$5y+440=105$$

Subtract 440 from both sides of the equation.

$$5y=-335$$

Divide both sides by 5.

$$y=-67$$

The system is now solved.

$$y=-67,x=-\frac{110}{3}$$