Solve {l}{2(2x-3)+3(y+4)=7}{4(x+2)-5(2-y)=-3}

$$\left\{ \begin{array} { l } { 2 ( 2 x - 3 ) + 3 ( y + 4 ) = 7 } \\ { 4 ( x + 2 ) - 5 ( 2 - y ) = - 3 } \end{array} \right.$$

Solve for x, y

$$x=1$$

$$y=-1$$

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.

$$2\left(2x-3\right)+3\left(y+4\right)=7,4\left(x+2\right)-5\left(-y+2\right)=-3$$

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

$$2\left(2x-3\right)+3\left(y+4\right)=7$$

Multiply 2 times 2x-3.

$$4x-6+3\left(y+4\right)=7$$

Multiply 3 times y+4.

$$4x-6+3y+12=7$$

Add -6 to 12.

$$4x+3y+6=7$$

Subtract 6 from both sides of the equation.

$$4x+3y=1$$

Subtract 3y from both sides of the equation.

$$4x=-3y+1$$

Divide both sides by 4.

$$x=\frac{1}{4}\left(-3y+1\right)$$

Multiply \frac{1}{4} times -3y+1.

$$x=-\frac{3}{4}y+\frac{1}{4}$$

Substitute \frac{-3y+1}{4} for x in the other equation, 4\left(x+2\right)-5\left(-y+2\right)=-3.

$$4\left(-\frac{3}{4}y+\frac{1}{4}+2\right)-5\left(-y+2\right)=-3$$

Add \frac{1}{4} to 2.

$$4\left(-\frac{3}{4}y+\frac{9}{4}\right)-5\left(-y+2\right)=-3$$

Multiply 4 times \frac{-3y+9}{4}.

$$-3y+9-5\left(-y+2\right)=-3$$

Multiply -5 times -y+2.

$$-3y+9+5y-10=-3$$

Add -3y to 5y.

$$2y+9-10=-3$$

Add 9 to -10.

$$2y-1=-3$$

Add 1 to both sides of the equation.

$$2y=-2$$

Divide both sides by 2.

$$y=-1$$

Substitute -1 for y in x=-\frac{3}{4}y+\frac{1}{4}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{3}{4}\left(-1\right)+\frac{1}{4}$$

Multiply -\frac{3}{4} times -1.

$$x=\frac{3+1}{4}$$

Add \frac{1}{4} to \frac{3}{4} 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.

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

Steps Using Matrices

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

$$2\left(2x-3\right)+3\left(y+4\right)=7,4\left(x+2\right)-5\left(-y+2\right)=-3$$

Simplify the first equation to put it in standard form.

$$2\left(2x-3\right)+3\left(y+4\right)=7$$

Multiply 2 times 2x-3.

$$4x-6+3\left(y+4\right)=7$$

Multiply 3 times y+4.

$$4x-6+3y+12=7$$

Add -6 to 12.

$$4x+3y+6=7$$

Subtract 6 from both sides of the equation.

$$4x+3y=1$$

Simplify the second equation to put it in standard form.

$$4\left(x+2\right)-5\left(-y+2\right)=-3$$

Multiply 4 times x+2.

$$4x+8-5\left(-y+2\right)=-3$$

Multiply -5 times -y+2.

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

Add 8 to -10.

$$4x+5y-2=-3$$

Add 2 to both sides of the equation.

$$4x+5y=-1$$

Write the equations in matrix form.

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

Left multiply the equation by the inverse matrix of \left(\begin{matrix}4&3\\4&5\end{matrix}\right).

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

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

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

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

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

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.

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

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{8}&-\frac{3}{8}\\-\frac{1}{2}&\frac{1}{2}\end{matrix}\right)\left(\begin{matrix}1\\-1\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{8}-\frac{3}{8}\left(-1\right)\\-\frac{1}{2}+\frac{1}{2}\left(-1\right)\end{matrix}\right)$$

Do the arithmetic.

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

Extract the matrix elements x and y.

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