Solve {l}{x/3+y/2=6}{2(3x-4)-3(y-1)=43}

$$\left\{ \begin{array} { l } { \frac { x } { 3 } + \frac { y } { 2 } = 6 } \\ { 2 ( 3 x - 4 ) - 3 ( y - 1 ) = 43 } \end{array} \right.$$

Solve for x, y

$$x = \frac{21}{2} = 10\frac{1}{2} = 10.5$$

$$y=5$$

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.

$$\frac{1}{3}x+\frac{1}{2}y=6,2\left(3x-4\right)-3\left(y-1\right)=43$$

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

$$\frac{1}{3}x+\frac{1}{2}y=6$$

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

$$\frac{1}{3}x=-\frac{1}{2}y+6$$

Multiply both sides by 3.

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

Multiply 3 times -\frac{y}{2}+6.

$$x=-\frac{3}{2}y+18$$

Substitute -\frac{3y}{2}+18 for x in the other equation, 2\left(3x-4\right)-3\left(y-1\right)=43.

$$2\left(3\left(-\frac{3}{2}y+18\right)-4\right)-3\left(y-1\right)=43$$

Multiply 3 times -\frac{3y}{2}+18.

$$2\left(-\frac{9}{2}y+54-4\right)-3\left(y-1\right)=43$$

Add 54 to -4.

$$2\left(-\frac{9}{2}y+50\right)-3\left(y-1\right)=43$$

Multiply 2 times -\frac{9y}{2}+50.

$$-9y+100-3\left(y-1\right)=43$$

Multiply -3 times y-1.

$$-9y+100-3y+3=43$$

Add -9y to -3y.

$$-12y+100+3=43$$

Add 100 to 3.

$$-12y+103=43$$

Subtract 103 from both sides of the equation.

$$-12y=-60$$

Divide both sides by -12.

$$y=5$$

Substitute 5 for y in x=-\frac{3}{2}y+18. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{3}{2}\times 5+18$$

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

$$x=-\frac{15}{2}+18$$

Add 18 to -\frac{15}{2}.

$$x=\frac{21}{2}$$

The system is now solved.

$$x=\frac{21}{2},y=5$$

Steps Using Matrices

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

$$\frac{1}{3}x+\frac{1}{2}y=6,2\left(3x-4\right)-3\left(y-1\right)=43$$

Simplify the second equation to put it in standard form.

$$2\left(3x-4\right)-3\left(y-1\right)=43$$

Multiply 2 times 3x-4.

$$6x-8-3\left(y-1\right)=43$$

Multiply -3 times y-1.

$$6x-8-3y+3=43$$

Add -8 to 3.

$$6x-3y-5=43$$

Add 5 to both sides of the equation.

$$6x-3y=48$$

Write the equations in matrix form.

$$\left(\begin{matrix}\frac{1}{3}&\frac{1}{2}\\6&-3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}6\\48\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}\frac{1}{3}&\frac{1}{2}\\6&-3\end{matrix}\right).

$$inverse(\left(\begin{matrix}\frac{1}{3}&\frac{1}{2}\\6&-3\end{matrix}\right))\left(\begin{matrix}\frac{1}{3}&\frac{1}{2}\\6&-3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{1}{3}&\frac{1}{2}\\6&-3\end{matrix}\right))\left(\begin{matrix}6\\48\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{1}{3}&\frac{1}{2}\\6&-3\end{matrix}\right))\left(\begin{matrix}6\\48\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{1}{3}&\frac{1}{2}\\6&-3\end{matrix}\right))\left(\begin{matrix}6\\48\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{3}{\frac{1}{3}\left(-3\right)-\frac{1}{2}\times 6}&-\frac{\frac{1}{2}}{\frac{1}{3}\left(-3\right)-\frac{1}{2}\times 6}\\-\frac{6}{\frac{1}{3}\left(-3\right)-\frac{1}{2}\times 6}&\frac{\frac{1}{3}}{\frac{1}{3}\left(-3\right)-\frac{1}{2}\times 6}\end{matrix}\right)\left(\begin{matrix}6\\48\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{4}&\frac{1}{8}\\\frac{3}{2}&-\frac{1}{12}\end{matrix}\right)\left(\begin{matrix}6\\48\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{4}\times 6+\frac{1}{8}\times 48\\\frac{3}{2}\times 6-\frac{1}{12}\times 48\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{21}{2}\\5\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{21}{2},y=5$$