Solve {l}{3x-2(y-1)=110}{x/4+y/3=3}

$$\left\{ \begin{array} { l } { 3 x - 2 ( y - 1 ) = 110 } \\ { \frac { x } { 4 } + \frac { y } { 3 } = 3 } \end{array} \right.$$

Solve for x, y

$$x=28$$

$$y=-12$$

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.

$$3x-2\left(y-1\right)=110,\frac{1}{4}x+\frac{1}{3}y=3$$

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

$$3x-2\left(y-1\right)=110$$

Multiply -2 times y-1.

$$3x-2y+2=110$$

Subtract 2 from both sides of the equation.

$$3x-2y=108$$

Add 2y to both sides of the equation.

$$3x=2y+108$$

Divide both sides by 3.

$$x=\frac{1}{3}\left(2y+108\right)$$

Multiply \frac{1}{3} times 108+2y.

$$x=\frac{2}{3}y+36$$

Substitute \frac{2y}{3}+36 for x in the other equation, \frac{1}{4}x+\frac{1}{3}y=3.

$$\frac{1}{4}\left(\frac{2}{3}y+36\right)+\frac{1}{3}y=3$$

Multiply \frac{1}{4} times \frac{2y}{3}+36.

$$\frac{1}{6}y+9+\frac{1}{3}y=3$$

Add \frac{y}{6} to \frac{y}{3}.

$$\frac{1}{2}y+9=3$$

Subtract 9 from both sides of the equation.

$$\frac{1}{2}y=-6$$

Multiply both sides by 2.

$$y=-12$$

Substitute -12 for y in x=\frac{2}{3}y+36. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{2}{3}\left(-12\right)+36$$

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

$$x=-8+36$$

Add 36 to -8.

$$x=28$$

The system is now solved.

$$x=28,y=-12$$

Steps Using Matrices

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

$$3x-2\left(y-1\right)=110,\frac{1}{4}x+\frac{1}{3}y=3$$

Simplify the first equation to put it in standard form.

$$3x-2\left(y-1\right)=110$$

Multiply -2 times y-1.

$$3x-2y+2=110$$

Subtract 2 from both sides of the equation.

$$3x-2y=108$$

Write the equations in matrix form.

$$\left(\begin{matrix}3&-2\\\frac{1}{4}&\frac{1}{3}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}108\\3\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}3&-2\\\frac{1}{4}&\frac{1}{3}\end{matrix}\right).

$$inverse(\left(\begin{matrix}3&-2\\\frac{1}{4}&\frac{1}{3}\end{matrix}\right))\left(\begin{matrix}3&-2\\\frac{1}{4}&\frac{1}{3}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&-2\\\frac{1}{4}&\frac{1}{3}\end{matrix}\right))\left(\begin{matrix}108\\3\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&-2\\\frac{1}{4}&\frac{1}{3}\end{matrix}\right))\left(\begin{matrix}108\\3\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&-2\\\frac{1}{4}&\frac{1}{3}\end{matrix}\right))\left(\begin{matrix}108\\3\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{\frac{1}{3}}{3\times \frac{1}{3}-\left(-2\times \frac{1}{4}\right)}&-\frac{-2}{3\times \frac{1}{3}-\left(-2\times \frac{1}{4}\right)}\\-\frac{\frac{1}{4}}{3\times \frac{1}{3}-\left(-2\times \frac{1}{4}\right)}&\frac{3}{3\times \frac{1}{3}-\left(-2\times \frac{1}{4}\right)}\end{matrix}\right)\left(\begin{matrix}108\\3\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{9}&\frac{4}{3}\\-\frac{1}{6}&2\end{matrix}\right)\left(\begin{matrix}108\\3\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{9}\times 108+\frac{4}{3}\times 3\\-\frac{1}{6}\times 108+2\times 3\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}28\\-12\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=28,y=-12$$