Solve {l}{-5a=4a+2b-3}{-b/2a=1}

$$\left\{ \begin{array} { l } { - 5 a = 4 a + 2 b - 3 } \\ { - \frac { b } { 2 a } = 1 } \end{array} \right.$$

Solve for a, b

$$a=\frac{3}{5}=0.6$$

$$b = -\frac{6}{5} = -1\frac{1}{5} = -1.2$$

Steps Using Substitution

Consider the first equation. Subtract 4a from both sides.

$$-5a-4a=2b-3$$

Combine -5a and -4a to get -9a.

$$-9a=2b-3$$

Divide both sides by -9.

$$a=-\frac{1}{9}\left(2b-3\right)$$

Multiply -\frac{1}{9} times 2b-3.

$$a=-\frac{2}{9}b+\frac{1}{3}$$

Substitute -\frac{2b}{9}+\frac{1}{3} for a in the other equation, -2a-b=0.

$$-2\left(-\frac{2}{9}b+\frac{1}{3}\right)-b=0$$

Multiply -2 times -\frac{2b}{9}+\frac{1}{3}.

$$\frac{4}{9}b-\frac{2}{3}-b=0$$

Add \frac{4b}{9} to -b.

$$-\frac{5}{9}b-\frac{2}{3}=0$$

Add \frac{2}{3} to both sides of the equation.

$$-\frac{5}{9}b=\frac{2}{3}$$

Divide both sides of the equation by -\frac{5}{9}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$b=-\frac{6}{5}$$

Substitute -\frac{6}{5} for b in a=-\frac{2}{9}b+\frac{1}{3}. Because the resulting equation contains only one variable, you can solve for a directly.

$$a=-\frac{2}{9}\left(-\frac{6}{5}\right)+\frac{1}{3}$$

Multiply -\frac{2}{9} times -\frac{6}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$a=\frac{4}{15}+\frac{1}{3}$$

Add \frac{1}{3} to \frac{4}{15} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$a=\frac{3}{5}$$

The system is now solved.

$$a=\frac{3}{5},b=-\frac{6}{5}$$

Steps Using Matrices

Consider the first equation. Subtract 4a from both sides.

$$-5a-4a=2b-3$$

Combine -5a and -4a to get -9a.

$$-9a=2b-3$$

Subtract 2b from both sides.

$$-9a-2b=-3$$

Consider the second equation. Variable a cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 2a.

$$-b=2a$$

Subtract 2a from both sides.

$$-b-2a=0$$

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

$$-9a-2b=-3,-2a-b=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}-9&-2\\-2&-1\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}-3\\0\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}-9&-2\\-2&-1\end{matrix}\right).

$$inverse(\left(\begin{matrix}-9&-2\\-2&-1\end{matrix}\right))\left(\begin{matrix}-9&-2\\-2&-1\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}-9&-2\\-2&-1\end{matrix}\right))\left(\begin{matrix}-3\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}-9&-2\\-2&-1\end{matrix}\right))\left(\begin{matrix}-3\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}-9&-2\\-2&-1\end{matrix}\right))\left(\begin{matrix}-3\\0\end{matrix}\right)$$

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.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{-9\left(-1\right)-\left(-2\left(-2\right)\right)}&-\frac{-2}{-9\left(-1\right)-\left(-2\left(-2\right)\right)}\\-\frac{-2}{-9\left(-1\right)-\left(-2\left(-2\right)\right)}&-\frac{9}{-9\left(-1\right)-\left(-2\left(-2\right)\right)}\end{matrix}\right)\left(\begin{matrix}-3\\0\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{5}&\frac{2}{5}\\\frac{2}{5}&-\frac{9}{5}\end{matrix}\right)\left(\begin{matrix}-3\\0\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{5}\left(-3\right)\\\frac{2}{5}\left(-3\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{3}{5}\\-\frac{6}{5}\end{matrix}\right)$$

Extract the matrix elements a and b.

$$a=\frac{3}{5},b=-\frac{6}{5}$$

Steps Using Elimination

Consider the first equation. Subtract 4a from both sides.

$$-5a-4a=2b-3$$

Combine -5a and -4a to get -9a.

$$-9a=2b-3$$

Subtract 2b from both sides.

$$-9a-2b=-3$$

Consider the second equation. Variable a cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 2a.

$$-b=2a$$

Subtract 2a from both sides.

$$-b-2a=0$$

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.

$$-9a-2b=-3,-2a-b=0$$

To make -9a and -2a equal, multiply all terms on each side of the first equation by -2 and all terms on each side of the second by -9.

$$-2\left(-9\right)a-2\left(-2\right)b=-2\left(-3\right),-9\left(-2\right)a-9\left(-1\right)b=0$$

Simplify.

$$18a+4b=6,18a+9b=0$$

Subtract 18a+9b=0 from 18a+4b=6 by subtracting like terms on each side of the equal sign.

$$18a-18a+4b-9b=6$$

Add 18a to -18a. Terms 18a and -18a cancel out, leaving an equation with only one variable that can be solved.

$$4b-9b=6$$

Add 4b to -9b.

$$-5b=6$$

Divide both sides by -5.

$$b=-\frac{6}{5}$$

Substitute -\frac{6}{5} for b in -2a-b=0. Because the resulting equation contains only one variable, you can solve for a directly.

$$-2a-\left(-\frac{6}{5}\right)=0$$

Subtract \frac{6}{5} from both sides of the equation.

$$-2a=-\frac{6}{5}$$

Divide both sides by -2.

$$a=\frac{3}{5}$$

The system is now solved.

$$a=\frac{3}{5},b=-\frac{6}{5}$$