Solve {l}{3x-4y=7}{x+3/2-y=4}

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

Solve for x, y

$$x=-3$$

$$y=-4$$

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.

$$3x-4y=7,\frac{1}{2}\left(x+3\right)-y=4$$

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

$$3x-4y=7$$

Add 4y to both sides of the equation.

$$3x=4y+7$$

Divide both sides by 3.

$$x=\frac{1}{3}\left(4y+7\right)$$

Multiply \frac{1}{3} times 4y+7.

$$x=\frac{4}{3}y+\frac{7}{3}$$

Substitute \frac{4y+7}{3} for x in the other equation, \frac{1}{2}\left(x+3\right)-y=4.

$$\frac{1}{2}\left(\frac{4}{3}y+\frac{7}{3}+3\right)-y=4$$

Add \frac{7}{3} to 3.

$$\frac{1}{2}\left(\frac{4}{3}y+\frac{16}{3}\right)-y=4$$

Multiply \frac{1}{2} times \frac{16+4y}{3}.

$$\frac{2}{3}y+\frac{8}{3}-y=4$$

Add \frac{2y}{3} to -y.

$$-\frac{1}{3}y+\frac{8}{3}=4$$

Subtract \frac{8}{3} from both sides of the equation.

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

Multiply both sides by -3.

$$y=-4$$

Substitute -4 for y in x=\frac{4}{3}y+\frac{7}{3}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{4}{3}\left(-4\right)+\frac{7}{3}$$

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

$$x=\frac{-16+7}{3}$$

Add \frac{7}{3} to -\frac{16}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-3$$

The system is now solved.

$$x=-3,y=-4$$

Steps Using Matrices

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

$$3x-4y=7,\frac{1}{2}\left(x+3\right)-y=4$$

Simplify the second equation to put it in standard form.

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

Multiply \frac{1}{2} times x+3.

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

Subtract \frac{3}{2} from both sides of the equation.

$$\frac{1}{2}x-y=\frac{5}{2}$$

Write the equations in matrix form.

$$\left(\begin{matrix}3&-4\\\frac{1}{2}&-1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}7\\\frac{5}{2}\end{matrix}\right)$$

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

$$inverse(\left(\begin{matrix}3&-4\\\frac{1}{2}&-1\end{matrix}\right))\left(\begin{matrix}3&-4\\\frac{1}{2}&-1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&-4\\\frac{1}{2}&-1\end{matrix}\right))\left(\begin{matrix}7\\\frac{5}{2}\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&-4\\\frac{1}{2}&-1\end{matrix}\right))\left(\begin{matrix}7\\\frac{5}{2}\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&-4\\\frac{1}{2}&-1\end{matrix}\right))\left(\begin{matrix}7\\\frac{5}{2}\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{3\left(-1\right)-\left(-4\times \frac{1}{2}\right)}&-\frac{-4}{3\left(-1\right)-\left(-4\times \frac{1}{2}\right)}\\-\frac{\frac{1}{2}}{3\left(-1\right)-\left(-4\times \frac{1}{2}\right)}&\frac{3}{3\left(-1\right)-\left(-4\times \frac{1}{2}\right)}\end{matrix}\right)\left(\begin{matrix}7\\\frac{5}{2}\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}1&-4\\\frac{1}{2}&-3\end{matrix}\right)\left(\begin{matrix}7\\\frac{5}{2}\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}7-4\times \frac{5}{2}\\\frac{1}{2}\times 7-3\times \frac{5}{2}\end{matrix}\right)$$

Do the arithmetic.

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

Extract the matrix elements x and y.

$$x=-3,y=-4$$