Solve {l}{h=2c-3}{201=3h+1.5c}

$$\left. \begin{array} { l } { h = 2 c - 3 } \\ { 201 = 3 h + 1.5 c } \end{array} \right.$$

Solve for h, c

$$h=53$$

$$c=28$$

Steps Using Substitution

Consider the first equation. Subtract 2c from both sides.

$$h-2c=-3$$

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

$$3h+1.5c=201$$

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.

$$h-2c=-3,3h+1.5c=201$$

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

$$h-2c=-3$$

Add 2c to both sides of the equation.

$$h=2c-3$$

Substitute 2c-3 for h in the other equation, 3h+1.5c=201.

$$3\left(2c-3\right)+1.5c=201$$

Multiply 3 times 2c-3.

$$6c-9+1.5c=201$$

Add 6c to \frac{3c}{2}.

$$7.5c-9=201$$

Add 9 to both sides of the equation.

$$7.5c=210$$

Divide both sides of the equation by 7.5, which is the same as multiplying both sides by the reciprocal of the fraction.

$$c=28$$

Substitute 28 for c in h=2c-3. Because the resulting equation contains only one variable, you can solve for h directly.

$$h=2\times 28-3$$

Multiply 2 times 28.

$$h=56-3$$

Add -3 to 56.

$$h=53$$

The system is now solved.

$$h=53,c=28$$

Steps Using Matrices

Consider the first equation. Subtract 2c from both sides.

$$h-2c=-3$$

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

$$3h+1.5c=201$$

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

$$h-2c=-3,3h+1.5c=201$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&-2\\3&1.5\end{matrix}\right)\left(\begin{matrix}h\\c\end{matrix}\right)=\left(\begin{matrix}-3\\201\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&-2\\3&1.5\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&-2\\3&1.5\end{matrix}\right))\left(\begin{matrix}1&-2\\3&1.5\end{matrix}\right)\left(\begin{matrix}h\\c\end{matrix}\right)=inverse(\left(\begin{matrix}1&-2\\3&1.5\end{matrix}\right))\left(\begin{matrix}-3\\201\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}h\\c\end{matrix}\right)=inverse(\left(\begin{matrix}1&-2\\3&1.5\end{matrix}\right))\left(\begin{matrix}-3\\201\end{matrix}\right)$$

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

$$\left(\begin{matrix}h\\c\end{matrix}\right)=inverse(\left(\begin{matrix}1&-2\\3&1.5\end{matrix}\right))\left(\begin{matrix}-3\\201\end{matrix}\right)$$

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.

$$\left(\begin{matrix}h\\c\end{matrix}\right)=\left(\begin{matrix}\frac{1.5}{1.5-\left(-2\times 3\right)}&-\frac{-2}{1.5-\left(-2\times 3\right)}\\-\frac{3}{1.5-\left(-2\times 3\right)}&\frac{1}{1.5-\left(-2\times 3\right)}\end{matrix}\right)\left(\begin{matrix}-3\\201\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}h\\c\end{matrix}\right)=\left(\begin{matrix}0.2&\frac{4}{15}\\-0.4&\frac{2}{15}\end{matrix}\right)\left(\begin{matrix}-3\\201\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}h\\c\end{matrix}\right)=\left(\begin{matrix}0.2\left(-3\right)+\frac{4}{15}\times 201\\-0.4\left(-3\right)+\frac{2}{15}\times 201\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}h\\c\end{matrix}\right)=\left(\begin{matrix}53\\28\end{matrix}\right)$$

Extract the matrix elements h and c.

$$h=53,c=28$$

Steps Using Elimination

Consider the first equation. Subtract 2c from both sides.

$$h-2c=-3$$

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

$$3h+1.5c=201$$

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.

$$h-2c=-3,3h+1.5c=201$$

To make h and 3h equal, multiply all terms on each side of the first equation by 3 and all terms on each side of the second by 1.

$$3h+3\left(-2\right)c=3\left(-3\right),3h+1.5c=201$$

Simplify.

$$3h-6c=-9,3h+1.5c=201$$

Subtract 3h+1.5c=201 from 3h-6c=-9 by subtracting like terms on each side of the equal sign.

$$3h-3h-6c-1.5c=-9-201$$

Add 3h to -3h. Terms 3h and -3h cancel out, leaving an equation with only one variable that can be solved.

$$-6c-1.5c=-9-201$$

Add -6c to -\frac{3c}{2}.

$$-7.5c=-9-201$$

Add -9 to -201.

$$-7.5c=-210$$

Divide both sides of the equation by -7.5, which is the same as multiplying both sides by the reciprocal of the fraction.

$$c=28$$

Substitute 28 for c in 3h+1.5c=201. Because the resulting equation contains only one variable, you can solve for h directly.

$$3h+1.5\times 28=201$$

Multiply 1.5 times 28.

$$3h+42=201$$

Subtract 42 from both sides of the equation.

$$3h=159$$

Divide both sides by 3.

$$h=53$$

The system is now solved.

$$h=53,c=28$$