Solve {l}{-11/10=-6-y/2-x}{7-y/-3-x=2/5}

$$\left. \begin{array} { l } { - \frac { 11 } { 10 } = \frac { - 6 - y } { 2 - x } } \\ { \frac { 7 - y } { - 3 - x } = \frac { 2 } { 5 } } \end{array} \right.$$

Solve for y, x

$$x=-8$$

$$y=5$$

Steps Using Substitution

Consider the first equation. Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by 10\left(x-2\right), the least common multiple of 10,2-x.

$$-11\left(x-2\right)=-10\left(-6-y\right)$$

Use the distributive property to multiply -11 by x-2.

$$-11x+22=-10\left(-6-y\right)$$

Use the distributive property to multiply -10 by -6-y.

$$-11x+22=60+10y$$

Subtract 10y from both sides.

$$-11x+22-10y=60$$

Subtract 22 from both sides.

$$-11x-10y=60-22$$

Subtract 22 from 60 to get 38.

$$-11x-10y=38$$

Consider the second equation. Variable x cannot be equal to -3 since division by zero is not defined. Multiply both sides of the equation by 5\left(x+3\right), the least common multiple of -3-x,5.

$$-5\left(7-y\right)=2\left(x+3\right)$$

Use the distributive property to multiply -5 by 7-y.

$$-35+5y=2\left(x+3\right)$$

Use the distributive property to multiply 2 by x+3.

$$-35+5y=2x+6$$

Subtract 2x from both sides.

$$-35+5y-2x=6$$

Add 35 to both sides.

$$5y-2x=6+35$$

Add 6 and 35 to get 41.

$$5y-2x=41$$

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.

$$-11x-10y=38,-2x+5y=41$$

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

$$-11x-10y=38$$

Add 10y to both sides of the equation.

$$-11x=10y+38$$

Divide both sides by -11.

$$x=-\frac{1}{11}\left(10y+38\right)$$

Multiply -\frac{1}{11} times 10y+38.

$$x=-\frac{10}{11}y-\frac{38}{11}$$

Substitute \frac{-10y-38}{11} for x in the other equation, -2x+5y=41.

$$-2\left(-\frac{10}{11}y-\frac{38}{11}\right)+5y=41$$

Multiply -2 times \frac{-10y-38}{11}.

$$\frac{20}{11}y+\frac{76}{11}+5y=41$$

Add \frac{20y}{11} to 5y.

$$\frac{75}{11}y+\frac{76}{11}=41$$

Subtract \frac{76}{11} from both sides of the equation.

$$\frac{75}{11}y=\frac{375}{11}$$

Divide both sides of the equation by \frac{75}{11}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=5$$

Substitute 5 for y in x=-\frac{10}{11}y-\frac{38}{11}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{10}{11}\times 5-\frac{38}{11}$$

Multiply -\frac{10}{11} times 5.

$$x=\frac{-50-38}{11}$$

Add -\frac{38}{11} to -\frac{50}{11} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-8$$

The system is now solved.

$$x=-8,y=5$$

Steps Using Matrices

Consider the first equation. Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by 10\left(x-2\right), the least common multiple of 10,2-x.

$$-11\left(x-2\right)=-10\left(-6-y\right)$$

Use the distributive property to multiply -11 by x-2.

$$-11x+22=-10\left(-6-y\right)$$

Use the distributive property to multiply -10 by -6-y.

$$-11x+22=60+10y$$

Subtract 10y from both sides.

$$-11x+22-10y=60$$

Subtract 22 from both sides.

$$-11x-10y=60-22$$

Subtract 22 from 60 to get 38.

$$-11x-10y=38$$

Consider the second equation. Variable x cannot be equal to -3 since division by zero is not defined. Multiply both sides of the equation by 5\left(x+3\right), the least common multiple of -3-x,5.

$$-5\left(7-y\right)=2\left(x+3\right)$$

Use the distributive property to multiply -5 by 7-y.

$$-35+5y=2\left(x+3\right)$$

Use the distributive property to multiply 2 by x+3.

$$-35+5y=2x+6$$

Subtract 2x from both sides.

$$-35+5y-2x=6$$

Add 35 to both sides.

$$5y-2x=6+35$$

Add 6 and 35 to get 41.

$$5y-2x=41$$

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

$$-11x-10y=38,-2x+5y=41$$

Write the equations in matrix form.

$$\left(\begin{matrix}-11&-10\\-2&5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}38\\41\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}-11&-10\\-2&5\end{matrix}\right).

$$inverse(\left(\begin{matrix}-11&-10\\-2&5\end{matrix}\right))\left(\begin{matrix}-11&-10\\-2&5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-11&-10\\-2&5\end{matrix}\right))\left(\begin{matrix}38\\41\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-11&-10\\-2&5\end{matrix}\right))\left(\begin{matrix}38\\41\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-11&-10\\-2&5\end{matrix}\right))\left(\begin{matrix}38\\41\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{-11\times 5-\left(-10\left(-2\right)\right)}&-\frac{-10}{-11\times 5-\left(-10\left(-2\right)\right)}\\-\frac{-2}{-11\times 5-\left(-10\left(-2\right)\right)}&-\frac{11}{-11\times 5-\left(-10\left(-2\right)\right)}\end{matrix}\right)\left(\begin{matrix}38\\41\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{15}&-\frac{2}{15}\\-\frac{2}{75}&\frac{11}{75}\end{matrix}\right)\left(\begin{matrix}38\\41\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{15}\times 38-\frac{2}{15}\times 41\\-\frac{2}{75}\times 38+\frac{11}{75}\times 41\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-8\\5\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=-8,y=5$$

Steps Using Elimination

Consider the first equation. Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by 10\left(x-2\right), the least common multiple of 10,2-x.

$$-11\left(x-2\right)=-10\left(-6-y\right)$$

Use the distributive property to multiply -11 by x-2.

$$-11x+22=-10\left(-6-y\right)$$

Use the distributive property to multiply -10 by -6-y.

$$-11x+22=60+10y$$

Subtract 10y from both sides.

$$-11x+22-10y=60$$

Subtract 22 from both sides.

$$-11x-10y=60-22$$

Subtract 22 from 60 to get 38.

$$-11x-10y=38$$

Consider the second equation. Variable x cannot be equal to -3 since division by zero is not defined. Multiply both sides of the equation by 5\left(x+3\right), the least common multiple of -3-x,5.

$$-5\left(7-y\right)=2\left(x+3\right)$$

Use the distributive property to multiply -5 by 7-y.

$$-35+5y=2\left(x+3\right)$$

Use the distributive property to multiply 2 by x+3.

$$-35+5y=2x+6$$

Subtract 2x from both sides.

$$-35+5y-2x=6$$

Add 35 to both sides.

$$5y-2x=6+35$$

Add 6 and 35 to get 41.

$$5y-2x=41$$

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.

$$-11x-10y=38,-2x+5y=41$$

To make -11x and -2x equal, multiply all terms on each side of the first equation by -2 and all terms on each side of the second by -11.

$$-2\left(-11\right)x-2\left(-10\right)y=-2\times 38,-11\left(-2\right)x-11\times 5y=-11\times 41$$

Simplify.

$$22x+20y=-76,22x-55y=-451$$

Subtract 22x-55y=-451 from 22x+20y=-76 by subtracting like terms on each side of the equal sign.

$$22x-22x+20y+55y=-76+451$$

Add 22x to -22x. Terms 22x and -22x cancel out, leaving an equation with only one variable that can be solved.

$$20y+55y=-76+451$$

Add 20y to 55y.

$$75y=-76+451$$

Add -76 to 451.

$$75y=375$$

Divide both sides by 75.

$$y=5$$

Substitute 5 for y in -2x+5y=41. Because the resulting equation contains only one variable, you can solve for x directly.

$$-2x+5\times 5=41$$

Multiply 5 times 5.

$$-2x+25=41$$

Subtract 25 from both sides of the equation.

$$-2x=16$$

Divide both sides by -2.

$$x=-8$$

The system is now solved.

$$x=-8,y=5$$