Solve {l}{5b-2h=124}{3b-4h=24}

$$\left\{ \begin{array} { l } { 5 b - 2 h = 124 } \\ { 3 b - 4 h = 24 } \end{array} \right.$$

Solve for b, h

$$b=32$$

$$h=18$$

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.

$$5b-2h=124,3b-4h=24$$

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

$$5b-2h=124$$

Add 2h to both sides of the equation.

$$5b=2h+124$$

Divide both sides by 5.

$$b=\frac{1}{5}\left(2h+124\right)$$

Multiply \frac{1}{5} times 124+2h.

$$b=\frac{2}{5}h+\frac{124}{5}$$

Substitute \frac{124+2h}{5} for b in the other equation, 3b-4h=24.

$$3\left(\frac{2}{5}h+\frac{124}{5}\right)-4h=24$$

Multiply 3 times \frac{124+2h}{5}.

$$\frac{6}{5}h+\frac{372}{5}-4h=24$$

Add \frac{6h}{5} to -4h.

$$-\frac{14}{5}h+\frac{372}{5}=24$$

Subtract \frac{372}{5} from both sides of the equation.

$$-\frac{14}{5}h=-\frac{252}{5}$$

Divide both sides of the equation by -\frac{14}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$h=18$$

Substitute 18 for h in b=\frac{2}{5}h+\frac{124}{5}. Because the resulting equation contains only one variable, you can solve for b directly.

$$b=\frac{2}{5}\times 18+\frac{124}{5}$$

Multiply \frac{2}{5} times 18.

$$b=\frac{36+124}{5}$$

Add \frac{124}{5} to \frac{36}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$b=32$$

The system is now solved.

$$b=32,h=18$$

Steps Using Matrices

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

$$5b-2h=124,3b-4h=24$$

Write the equations in matrix form.

$$\left(\begin{matrix}5&-2\\3&-4\end{matrix}\right)\left(\begin{matrix}b\\h\end{matrix}\right)=\left(\begin{matrix}124\\24\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}5&-2\\3&-4\end{matrix}\right).

$$inverse(\left(\begin{matrix}5&-2\\3&-4\end{matrix}\right))\left(\begin{matrix}5&-2\\3&-4\end{matrix}\right)\left(\begin{matrix}b\\h\end{matrix}\right)=inverse(\left(\begin{matrix}5&-2\\3&-4\end{matrix}\right))\left(\begin{matrix}124\\24\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}b\\h\end{matrix}\right)=inverse(\left(\begin{matrix}5&-2\\3&-4\end{matrix}\right))\left(\begin{matrix}124\\24\end{matrix}\right)$$

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

$$\left(\begin{matrix}b\\h\end{matrix}\right)=inverse(\left(\begin{matrix}5&-2\\3&-4\end{matrix}\right))\left(\begin{matrix}124\\24\end{matrix}\right)$$

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.

$$\left(\begin{matrix}b\\h\end{matrix}\right)=\left(\begin{matrix}-\frac{4}{5\left(-4\right)-\left(-2\times 3\right)}&-\frac{-2}{5\left(-4\right)-\left(-2\times 3\right)}\\-\frac{3}{5\left(-4\right)-\left(-2\times 3\right)}&\frac{5}{5\left(-4\right)-\left(-2\times 3\right)}\end{matrix}\right)\left(\begin{matrix}124\\24\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}b\\h\end{matrix}\right)=\left(\begin{matrix}\frac{2}{7}&-\frac{1}{7}\\\frac{3}{14}&-\frac{5}{14}\end{matrix}\right)\left(\begin{matrix}124\\24\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}b\\h\end{matrix}\right)=\left(\begin{matrix}\frac{2}{7}\times 124-\frac{1}{7}\times 24\\\frac{3}{14}\times 124-\frac{5}{14}\times 24\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}b\\h\end{matrix}\right)=\left(\begin{matrix}32\\18\end{matrix}\right)$$

Extract the matrix elements b and h.

$$b=32,h=18$$

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.

$$5b-2h=124,3b-4h=24$$

To make 5b and 3b equal, multiply all terms on each side of the first equation by 3 and all terms on each side of the second by 5.

$$3\times 5b+3\left(-2\right)h=3\times 124,5\times 3b+5\left(-4\right)h=5\times 24$$

Simplify.

$$15b-6h=372,15b-20h=120$$

Subtract 15b-20h=120 from 15b-6h=372 by subtracting like terms on each side of the equal sign.

$$15b-15b-6h+20h=372-120$$

Add 15b to -15b. Terms 15b and -15b cancel out, leaving an equation with only one variable that can be solved.

$$-6h+20h=372-120$$

Add -6h to 20h.

$$14h=372-120$$

Add 372 to -120.

$$14h=252$$

Divide both sides by 14.

$$h=18$$

Substitute 18 for h in 3b-4h=24. Because the resulting equation contains only one variable, you can solve for b directly.

$$3b-4\times 18=24$$

Multiply -4 times 18.

$$3b-72=24$$

Add 72 to both sides of the equation.

$$3b=96$$

Divide both sides by 3.

$$b=32$$

The system is now solved.

$$b=32,h=18$$