Solve {l}{20a+3b=58}{15a+4b=47}

$$\left. \begin{array} { l } { 20 a + 3 b = 58 } \\ { 15 a + 4 b = 47 } \end{array} \right.$$

Solve for a, b

$$a = \frac{13}{5} = 2\frac{3}{5} = 2.6$$

$$b=2$$

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.

$$20a+3b=58,15a+4b=47$$

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

$$20a+3b=58$$

Subtract 3b from both sides of the equation.

$$20a=-3b+58$$

Divide both sides by 20.

$$a=\frac{1}{20}\left(-3b+58\right)$$

Multiply \frac{1}{20} times -3b+58.

$$a=-\frac{3}{20}b+\frac{29}{10}$$

Substitute -\frac{3b}{20}+\frac{29}{10} for a in the other equation, 15a+4b=47.

$$15\left(-\frac{3}{20}b+\frac{29}{10}\right)+4b=47$$

Multiply 15 times -\frac{3b}{20}+\frac{29}{10}.

$$-\frac{9}{4}b+\frac{87}{2}+4b=47$$

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

$$\frac{7}{4}b+\frac{87}{2}=47$$

Subtract \frac{87}{2} from both sides of the equation.

$$\frac{7}{4}b=\frac{7}{2}$$

Divide both sides of the equation by \frac{7}{4}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$b=2$$

Substitute 2 for b in a=-\frac{3}{20}b+\frac{29}{10}. Because the resulting equation contains only one variable, you can solve for a directly.

$$a=-\frac{3}{20}\times 2+\frac{29}{10}$$

Multiply -\frac{3}{20} times 2.

$$a=\frac{-3+29}{10}$$

Add \frac{29}{10} to -\frac{3}{10} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$a=\frac{13}{5}$$

The system is now solved.

$$a=\frac{13}{5},b=2$$

Steps Using Matrices

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

$$20a+3b=58,15a+4b=47$$

Write the equations in matrix form.

$$\left(\begin{matrix}20&3\\15&4\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}58\\47\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}20&3\\15&4\end{matrix}\right).

$$inverse(\left(\begin{matrix}20&3\\15&4\end{matrix}\right))\left(\begin{matrix}20&3\\15&4\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}20&3\\15&4\end{matrix}\right))\left(\begin{matrix}58\\47\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}20&3\\15&4\end{matrix}\right))\left(\begin{matrix}58\\47\end{matrix}\right)$$

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

$$\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}20&3\\15&4\end{matrix}\right))\left(\begin{matrix}58\\47\end{matrix}\right)$$

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.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{4}{20\times 4-3\times 15}&-\frac{3}{20\times 4-3\times 15}\\-\frac{15}{20\times 4-3\times 15}&\frac{20}{20\times 4-3\times 15}\end{matrix}\right)\left(\begin{matrix}58\\47\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{4}{35}&-\frac{3}{35}\\-\frac{3}{7}&\frac{4}{7}\end{matrix}\right)\left(\begin{matrix}58\\47\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{4}{35}\times 58-\frac{3}{35}\times 47\\-\frac{3}{7}\times 58+\frac{4}{7}\times 47\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{13}{5}\\2\end{matrix}\right)$$

Extract the matrix elements a and b.

$$a=\frac{13}{5},b=2$$

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.

$$20a+3b=58,15a+4b=47$$

To make 20a and 15a equal, multiply all terms on each side of the first equation by 15 and all terms on each side of the second by 20.

$$15\times 20a+15\times 3b=15\times 58,20\times 15a+20\times 4b=20\times 47$$

Simplify.

$$300a+45b=870,300a+80b=940$$

Subtract 300a+80b=940 from 300a+45b=870 by subtracting like terms on each side of the equal sign.

$$300a-300a+45b-80b=870-940$$

Add 300a to -300a. Terms 300a and -300a cancel out, leaving an equation with only one variable that can be solved.

$$45b-80b=870-940$$

Add 45b to -80b.

$$-35b=870-940$$

Add 870 to -940.

$$-35b=-70$$

Divide both sides by -35.

$$b=2$$

Substitute 2 for b in 15a+4b=47. Because the resulting equation contains only one variable, you can solve for a directly.

$$15a+4\times 2=47$$

Multiply 4 times 2.

$$15a+8=47$$

Subtract 8 from both sides of the equation.

$$15a=39$$

Divide both sides by 15.

$$a=\frac{13}{5}$$

The system is now solved.

$$a=\frac{13}{5},b=2$$