Solve {l}{0=a-b+5/2}{5/2=16a+4b+5/2}

$$\left\{ \begin{array} { l } { 0 = a - b + \frac { 5 } { 2 } } \\ { \frac { 5 } { 2 } = 16 a + 4 b + \frac { 5 } { 2 } } \end{array} \right.$$

Solve for a, b

$$a=-\frac{1}{2}=-0.5$$

$$b=2$$

Steps Using Substitution

Consider the first equation. Swap sides so that all variable terms are on the left hand side.

$$a-b+\frac{5}{2}=0$$

Subtract \frac{5}{2} from both sides. Anything subtracted from zero gives its negation.

$$a-b=-\frac{5}{2}$$

Consider the second equation. Swap sides so that all variable terms are on the left hand side.

$$16a+4b+\frac{5}{2}=\frac{5}{2}$$

Subtract \frac{5}{2} from both sides.

$$16a+4b=\frac{5}{2}-\frac{5}{2}$$

Subtract \frac{5}{2} from \frac{5}{2} to get 0.

$$16a+4b=0$$

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.

$$a-b=-\frac{5}{2},16a+4b=0$$

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

$$a-b=-\frac{5}{2}$$

Add b to both sides of the equation.

$$a=b-\frac{5}{2}$$

Substitute b-\frac{5}{2} for a in the other equation, 16a+4b=0.

$$16\left(b-\frac{5}{2}\right)+4b=0$$

Multiply 16 times b-\frac{5}{2}.

$$16b-40+4b=0$$

Add 16b to 4b.

$$20b-40=0$$

Add 40 to both sides of the equation.

$$20b=40$$

Divide both sides by 20.

$$b=2$$

Substitute 2 for b in a=b-\frac{5}{2}. Because the resulting equation contains only one variable, you can solve for a directly.

$$a=2-\frac{5}{2}$$

Add -\frac{5}{2} to 2.

$$a=-\frac{1}{2}$$

The system is now solved.

$$a=-\frac{1}{2},b=2$$

Steps Using Matrices

Consider the first equation. Swap sides so that all variable terms are on the left hand side.

$$a-b+\frac{5}{2}=0$$

Subtract \frac{5}{2} from both sides. Anything subtracted from zero gives its negation.

$$a-b=-\frac{5}{2}$$

Consider the second equation. Swap sides so that all variable terms are on the left hand side.

$$16a+4b+\frac{5}{2}=\frac{5}{2}$$

Subtract \frac{5}{2} from both sides.

$$16a+4b=\frac{5}{2}-\frac{5}{2}$$

Subtract \frac{5}{2} from \frac{5}{2} to get 0.

$$16a+4b=0$$

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

$$a-b=-\frac{5}{2},16a+4b=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&-1\\16&4\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}-\frac{5}{2}\\0\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&-1\\16&4\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&-1\\16&4\end{matrix}\right))\left(\begin{matrix}1&-1\\16&4\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}1&-1\\16&4\end{matrix}\right))\left(\begin{matrix}-\frac{5}{2}\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}1&-1\\16&4\end{matrix}\right))\left(\begin{matrix}-\frac{5}{2}\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}1&-1\\16&4\end{matrix}\right))\left(\begin{matrix}-\frac{5}{2}\\0\end{matrix}\right)$$

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.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{4}{4-\left(-16\right)}&-\frac{-1}{4-\left(-16\right)}\\-\frac{16}{4-\left(-16\right)}&\frac{1}{4-\left(-16\right)}\end{matrix}\right)\left(\begin{matrix}-\frac{5}{2}\\0\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{1}{5}&\frac{1}{20}\\-\frac{4}{5}&\frac{1}{20}\end{matrix}\right)\left(\begin{matrix}-\frac{5}{2}\\0\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{1}{5}\left(-\frac{5}{2}\right)\\-\frac{4}{5}\left(-\frac{5}{2}\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{2}\\2\end{matrix}\right)$$

Extract the matrix elements a and b.

$$a=-\frac{1}{2},b=2$$

Steps Using Elimination

Consider the first equation. Swap sides so that all variable terms are on the left hand side.

$$a-b+\frac{5}{2}=0$$

Subtract \frac{5}{2} from both sides. Anything subtracted from zero gives its negation.

$$a-b=-\frac{5}{2}$$

Consider the second equation. Swap sides so that all variable terms are on the left hand side.

$$16a+4b+\frac{5}{2}=\frac{5}{2}$$

Subtract \frac{5}{2} from both sides.

$$16a+4b=\frac{5}{2}-\frac{5}{2}$$

Subtract \frac{5}{2} from \frac{5}{2} to get 0.

$$16a+4b=0$$

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.

$$a-b=-\frac{5}{2},16a+4b=0$$

To make a and 16a equal, multiply all terms on each side of the first equation by 16 and all terms on each side of the second by 1.

$$16a+16\left(-1\right)b=16\left(-\frac{5}{2}\right),16a+4b=0$$

Simplify.

$$16a-16b=-40,16a+4b=0$$

Subtract 16a+4b=0 from 16a-16b=-40 by subtracting like terms on each side of the equal sign.

$$16a-16a-16b-4b=-40$$

Add 16a to -16a. Terms 16a and -16a cancel out, leaving an equation with only one variable that can be solved.

$$-16b-4b=-40$$

Add -16b to -4b.

$$-20b=-40$$

Divide both sides by -20.

$$b=2$$

Substitute 2 for b in 16a+4b=0. Because the resulting equation contains only one variable, you can solve for a directly.

$$16a+4\times 2=0$$

Multiply 4 times 2.

$$16a+8=0$$

Subtract 8 from both sides of the equation.

$$16a=-8$$

Divide both sides by 16.

$$a=-\frac{1}{2}$$

The system is now solved.

$$a=-\frac{1}{2},b=2$$