Solve {l}{x+3y-2(1+y-2x)=-9}{4x-3y-5(1+y-2x)=-3}

$$\left. \begin{array} { l } { x + 3 y - 2 ( 1 + y - 2 x ) = - 9 } \\ { 4 x - 3 y - 5 ( 1 + y - 2 x ) = - 3 } \end{array} \right.$$

Solve for x, y

$$x=-1$$

$$y=-2$$

Steps Using Substitution

Consider the first equation. Use the distributive property to multiply -2 by 1+y-2x.

$$x+3y-2-2y+4x=-9$$

Combine 3y and -2y to get y.

$$x+y-2+4x=-9$$

Combine x and 4x to get 5x.

$$5x+y-2=-9$$

Add 2 to both sides.

$$5x+y=-9+2$$

Add -9 and 2 to get -7.

$$5x+y=-7$$

Consider the second equation. Use the distributive property to multiply -5 by 1+y-2x.

$$4x-3y-5-5y+10x=-3$$

Combine -3y and -5y to get -8y.

$$4x-8y-5+10x=-3$$

Combine 4x and 10x to get 14x.

$$14x-8y-5=-3$$

Add 5 to both sides.

$$14x-8y=-3+5$$

Add -3 and 5 to get 2.

$$14x-8y=2$$

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.

$$5x+y=-7,14x-8y=2$$

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

$$5x+y=-7$$

Subtract y from both sides of the equation.

$$5x=-y-7$$

Divide both sides by 5.

$$x=\frac{1}{5}\left(-y-7\right)$$

Multiply \frac{1}{5} times -y-7.

$$x=-\frac{1}{5}y-\frac{7}{5}$$

Substitute \frac{-y-7}{5} for x in the other equation, 14x-8y=2.

$$14\left(-\frac{1}{5}y-\frac{7}{5}\right)-8y=2$$

Multiply 14 times \frac{-y-7}{5}.

$$-\frac{14}{5}y-\frac{98}{5}-8y=2$$

Add -\frac{14y}{5} to -8y.

$$-\frac{54}{5}y-\frac{98}{5}=2$$

Add \frac{98}{5} to both sides of the equation.

$$-\frac{54}{5}y=\frac{108}{5}$$

Divide both sides of the equation by -\frac{54}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=-2$$

Substitute -2 for y in x=-\frac{1}{5}y-\frac{7}{5}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{1}{5}\left(-2\right)-\frac{7}{5}$$

Multiply -\frac{1}{5} times -2.

$$x=\frac{2-7}{5}$$

Add -\frac{7}{5} to \frac{2}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-1$$

The system is now solved.

$$x=-1,y=-2$$

Steps Using Matrices

Consider the first equation. Use the distributive property to multiply -2 by 1+y-2x.

$$x+3y-2-2y+4x=-9$$

Combine 3y and -2y to get y.

$$x+y-2+4x=-9$$

Combine x and 4x to get 5x.

$$5x+y-2=-9$$

Add 2 to both sides.

$$5x+y=-9+2$$

Add -9 and 2 to get -7.

$$5x+y=-7$$

Consider the second equation. Use the distributive property to multiply -5 by 1+y-2x.

$$4x-3y-5-5y+10x=-3$$

Combine -3y and -5y to get -8y.

$$4x-8y-5+10x=-3$$

Combine 4x and 10x to get 14x.

$$14x-8y-5=-3$$

Add 5 to both sides.

$$14x-8y=-3+5$$

Add -3 and 5 to get 2.

$$14x-8y=2$$

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

$$5x+y=-7,14x-8y=2$$

Write the equations in matrix form.

$$\left(\begin{matrix}5&1\\14&-8\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-7\\2\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}5&1\\14&-8\end{matrix}\right).

$$inverse(\left(\begin{matrix}5&1\\14&-8\end{matrix}\right))\left(\begin{matrix}5&1\\14&-8\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&1\\14&-8\end{matrix}\right))\left(\begin{matrix}-7\\2\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&1\\14&-8\end{matrix}\right))\left(\begin{matrix}-7\\2\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&1\\14&-8\end{matrix}\right))\left(\begin{matrix}-7\\2\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{8}{5\left(-8\right)-14}&-\frac{1}{5\left(-8\right)-14}\\-\frac{14}{5\left(-8\right)-14}&\frac{5}{5\left(-8\right)-14}\end{matrix}\right)\left(\begin{matrix}-7\\2\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{4}{27}&\frac{1}{54}\\\frac{7}{27}&-\frac{5}{54}\end{matrix}\right)\left(\begin{matrix}-7\\2\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{4}{27}\left(-7\right)+\frac{1}{54}\times 2\\\frac{7}{27}\left(-7\right)-\frac{5}{54}\times 2\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-1\\-2\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=-1,y=-2$$

Steps Using Elimination

Consider the first equation. Use the distributive property to multiply -2 by 1+y-2x.

$$x+3y-2-2y+4x=-9$$

Combine 3y and -2y to get y.

$$x+y-2+4x=-9$$

Combine x and 4x to get 5x.

$$5x+y-2=-9$$

Add 2 to both sides.

$$5x+y=-9+2$$

Add -9 and 2 to get -7.

$$5x+y=-7$$

Consider the second equation. Use the distributive property to multiply -5 by 1+y-2x.

$$4x-3y-5-5y+10x=-3$$

Combine -3y and -5y to get -8y.

$$4x-8y-5+10x=-3$$

Combine 4x and 10x to get 14x.

$$14x-8y-5=-3$$

Add 5 to both sides.

$$14x-8y=-3+5$$

Add -3 and 5 to get 2.

$$14x-8y=2$$

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.

$$5x+y=-7,14x-8y=2$$

To make 5x and 14x equal, multiply all terms on each side of the first equation by 14 and all terms on each side of the second by 5.

$$14\times 5x+14y=14\left(-7\right),5\times 14x+5\left(-8\right)y=5\times 2$$

Simplify.

$$70x+14y=-98,70x-40y=10$$

Subtract 70x-40y=10 from 70x+14y=-98 by subtracting like terms on each side of the equal sign.

$$70x-70x+14y+40y=-98-10$$

Add 70x to -70x. Terms 70x and -70x cancel out, leaving an equation with only one variable that can be solved.

$$14y+40y=-98-10$$

Add 14y to 40y.

$$54y=-98-10$$

Add -98 to -10.

$$54y=-108$$

Divide both sides by 54.

$$y=-2$$

Substitute -2 for y in 14x-8y=2. Because the resulting equation contains only one variable, you can solve for x directly.

$$14x-8\left(-2\right)=2$$

Multiply -8 times -2.

$$14x+16=2$$

Subtract 16 from both sides of the equation.

$$14x=-14$$

Divide both sides by 14.

$$x=-1$$

The system is now solved.

$$x=-1,y=-2$$