Solve {l}{2u+3v=8}{3u+4v=-3}

$$\left. \begin{array} { l } { 2 u + 3 v = 8 } \\ { 3 u + 4 v = - 3 } \end{array} \right.$$

Solve for u, v

$$u=-41$$

$$v=30$$

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.

$$2u+3v=8,3u+4v=-3$$

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

$$2u+3v=8$$

Subtract 3v from both sides of the equation.

$$2u=-3v+8$$

Divide both sides by 2.

$$u=\frac{1}{2}\left(-3v+8\right)$$

Multiply \frac{1}{2} times -3v+8.

$$u=-\frac{3}{2}v+4$$

Substitute -\frac{3v}{2}+4 for u in the other equation, 3u+4v=-3.

$$3\left(-\frac{3}{2}v+4\right)+4v=-3$$

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

$$-\frac{9}{2}v+12+4v=-3$$

Add -\frac{9v}{2} to 4v.

$$-\frac{1}{2}v+12=-3$$

Subtract 12 from both sides of the equation.

$$-\frac{1}{2}v=-15$$

Multiply both sides by -2.

$$v=30$$

Substitute 30 for v in u=-\frac{3}{2}v+4. Because the resulting equation contains only one variable, you can solve for u directly.

$$u=-\frac{3}{2}\times 30+4$$

Multiply -\frac{3}{2} times 30.

$$u=-45+4$$

Add 4 to -45.

$$u=-41$$

The system is now solved.

$$u=-41,v=30$$

Steps Using Matrices

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

$$2u+3v=8,3u+4v=-3$$

Write the equations in matrix form.

$$\left(\begin{matrix}2&3\\3&4\end{matrix}\right)\left(\begin{matrix}u\\v\end{matrix}\right)=\left(\begin{matrix}8\\-3\end{matrix}\right)$$

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

$$inverse(\left(\begin{matrix}2&3\\3&4\end{matrix}\right))\left(\begin{matrix}2&3\\3&4\end{matrix}\right)\left(\begin{matrix}u\\v\end{matrix}\right)=inverse(\left(\begin{matrix}2&3\\3&4\end{matrix}\right))\left(\begin{matrix}8\\-3\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}u\\v\end{matrix}\right)=inverse(\left(\begin{matrix}2&3\\3&4\end{matrix}\right))\left(\begin{matrix}8\\-3\end{matrix}\right)$$

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

$$\left(\begin{matrix}u\\v\end{matrix}\right)=inverse(\left(\begin{matrix}2&3\\3&4\end{matrix}\right))\left(\begin{matrix}8\\-3\end{matrix}\right)$$

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.

$$\left(\begin{matrix}u\\v\end{matrix}\right)=\left(\begin{matrix}\frac{4}{2\times 4-3\times 3}&-\frac{3}{2\times 4-3\times 3}\\-\frac{3}{2\times 4-3\times 3}&\frac{2}{2\times 4-3\times 3}\end{matrix}\right)\left(\begin{matrix}8\\-3\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}u\\v\end{matrix}\right)=\left(\begin{matrix}-4&3\\3&-2\end{matrix}\right)\left(\begin{matrix}8\\-3\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}u\\v\end{matrix}\right)=\left(\begin{matrix}-4\times 8+3\left(-3\right)\\3\times 8-2\left(-3\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}u\\v\end{matrix}\right)=\left(\begin{matrix}-41\\30\end{matrix}\right)$$

Extract the matrix elements u and v.

$$u=-41,v=30$$

Steps Using Elimination

In order to solve by elimination, coefficients of one of the variables must be the same in both equations so that the variable will cancel out when one equation is subtracted from the other.

$$2u+3v=8,3u+4v=-3$$

To make 2u and 3u equal, multiply all terms on each side of the first equation by 3 and all terms on each side of the second by 2.

$$3\times 2u+3\times 3v=3\times 8,2\times 3u+2\times 4v=2\left(-3\right)$$

Simplify.

$$6u+9v=24,6u+8v=-6$$

Subtract 6u+8v=-6 from 6u+9v=24 by subtracting like terms on each side of the equal sign.

$$6u-6u+9v-8v=24+6$$

Add 6u to -6u. Terms 6u and -6u cancel out, leaving an equation with only one variable that can be solved.

$$9v-8v=24+6$$

Add 9v to -8v.

$$v=24+6$$

Add 24 to 6.

$$v=30$$

Substitute 30 for v in 3u+4v=-3. Because the resulting equation contains only one variable, you can solve for u directly.

$$3u+4\times 30=-3$$

Multiply 4 times 30.

$$3u+120=-3$$

Subtract 120 from both sides of the equation.

$$3u=-123$$

Divide both sides by 3.

$$u=-41$$

The system is now solved.

$$u=-41,v=30$$