Solve {l}{4(x-1)+3(y+2)=4}{-5(x-2)-4(y-1)=13}

$$\left\{ \begin{array} { l } { 4 ( x - 1 ) + 3 ( y + 2 ) = 4 } \\ { - 5 ( x - 2 ) - 4 ( y - 1 ) = 13 } \end{array} \right.$$

Solve for x, y

$$x=5$$

$$y=-6$$

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.

$$4\left(x-1\right)+3\left(y+2\right)=4,-5\left(x-2\right)-4\left(y-1\right)=13$$

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

$$4\left(x-1\right)+3\left(y+2\right)=4$$

Multiply 4 times x-1.

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

Multiply 3 times y+2.

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

Add -4 to 6.

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

Subtract 2 from both sides of the equation.

$$4x+3y=2$$

Subtract 3y from both sides of the equation.

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

Divide both sides by 4.

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

Multiply \frac{1}{4} times -3y+2.

$$x=-\frac{3}{4}y+\frac{1}{2}$$

Substitute -\frac{3y}{4}+\frac{1}{2} for x in the other equation, -5\left(x-2\right)-4\left(y-1\right)=13.

$$-5\left(-\frac{3}{4}y+\frac{1}{2}-2\right)-4\left(y-1\right)=13$$

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

$$-5\left(-\frac{3}{4}y-\frac{3}{2}\right)-4\left(y-1\right)=13$$

Multiply -5 times -\frac{3y}{4}-\frac{3}{2}.

$$\frac{15}{4}y+\frac{15}{2}-4\left(y-1\right)=13$$

Multiply -4 times y-1.

$$\frac{15}{4}y+\frac{15}{2}-4y+4=13$$

Add \frac{15y}{4} to -4y.

$$-\frac{1}{4}y+\frac{15}{2}+4=13$$

Add \frac{15}{2} to 4.

$$-\frac{1}{4}y+\frac{23}{2}=13$$

Subtract \frac{23}{2} from both sides of the equation.

$$-\frac{1}{4}y=\frac{3}{2}$$

Multiply both sides by -4.

$$y=-6$$

Substitute -6 for y in x=-\frac{3}{4}y+\frac{1}{2}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{3}{4}\left(-6\right)+\frac{1}{2}$$

Multiply -\frac{3}{4} times -6.

$$x=\frac{9+1}{2}$$

Add \frac{1}{2} to \frac{9}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=5$$

The system is now solved.

$$x=5,y=-6$$

Steps Using Matrices

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

$$4\left(x-1\right)+3\left(y+2\right)=4,-5\left(x-2\right)-4\left(y-1\right)=13$$

Simplify the first equation to put it in standard form.

$$4\left(x-1\right)+3\left(y+2\right)=4$$

Multiply 4 times x-1.

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

Multiply 3 times y+2.

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

Add -4 to 6.

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

Subtract 2 from both sides of the equation.

$$4x+3y=2$$

Simplify the second equation to put it in standard form.

$$-5\left(x-2\right)-4\left(y-1\right)=13$$

Multiply -5 times x-2.

$$-5x+10-4\left(y-1\right)=13$$

Multiply -4 times y-1.

$$-5x+10-4y+4=13$$

Add 10 to 4.

$$-5x-4y+14=13$$

Subtract 14 from both sides of the equation.

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

Write the equations in matrix form.

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

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

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

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

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

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

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

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.

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

Do the arithmetic.

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

Multiply the matrices.

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

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}5\\-6\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=5,y=-6$$