Solve {l}{-10=c_{1}+c_{2}}{20=2c_{1}+4c_{2}}

$$\left\{ \begin{array} { l } { - 10 = c _ { 1 } + c _ { 2 } } \\ { 20 = 2 c _ { 1 } + 4 c _ { 2 } } \end{array} \right.$$

Solve for c_1, c_2

$$c_{1}=-30$$

$$c_{2}=20$$

Steps Using Substitution

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

$$c_{1}+c_{2}=-10$$

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

$$2c_{1}+4c_{2}=20$$

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.

$$c_{1}+c_{2}=-10,2c_{1}+4c_{2}=20$$

Choose one of the equations and solve it for c_{1} by isolating c_{1} on the left hand side of the equal sign.

$$c_{1}+c_{2}=-10$$

Subtract c_{2} from both sides of the equation.

$$c_{1}=-c_{2}-10$$

Substitute -c_{2}-10 for c_{1} in the other equation, 2c_{1}+4c_{2}=20.

$$2\left(-c_{2}-10\right)+4c_{2}=20$$

Multiply 2 times -c_{2}-10.

$$-2c_{2}-20+4c_{2}=20$$

Add -2c_{2} to 4c_{2}.

$$2c_{2}-20=20$$

Add 20 to both sides of the equation.

$$2c_{2}=40$$

Divide both sides by 2.

$$c_{2}=20$$

Substitute 20 for c_{2} in c_{1}=-c_{2}-10. Because the resulting equation contains only one variable, you can solve for c_{1} directly.

$$c_{1}=-20-10$$

Add -10 to -20.

$$c_{1}=-30$$

The system is now solved.

$$c_{1}=-30,c_{2}=20$$

Steps Using Matrices

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

$$c_{1}+c_{2}=-10$$

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

$$2c_{1}+4c_{2}=20$$

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

$$c_{1}+c_{2}=-10,2c_{1}+4c_{2}=20$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&1\\2&4\end{matrix}\right)\left(\begin{matrix}c_{1}\\c_{2}\end{matrix}\right)=\left(\begin{matrix}-10\\20\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\2&4\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&1\\2&4\end{matrix}\right))\left(\begin{matrix}1&1\\2&4\end{matrix}\right)\left(\begin{matrix}c_{1}\\c_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\2&4\end{matrix}\right))\left(\begin{matrix}-10\\20\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}c_{1}\\c_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\2&4\end{matrix}\right))\left(\begin{matrix}-10\\20\end{matrix}\right)$$

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

$$\left(\begin{matrix}c_{1}\\c_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\2&4\end{matrix}\right))\left(\begin{matrix}-10\\20\end{matrix}\right)$$

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.

$$\left(\begin{matrix}c_{1}\\c_{2}\end{matrix}\right)=\left(\begin{matrix}\frac{4}{4-2}&-\frac{1}{4-2}\\-\frac{2}{4-2}&\frac{1}{4-2}\end{matrix}\right)\left(\begin{matrix}-10\\20\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}c_{1}\\c_{2}\end{matrix}\right)=\left(\begin{matrix}2&-\frac{1}{2}\\-1&\frac{1}{2}\end{matrix}\right)\left(\begin{matrix}-10\\20\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}c_{1}\\c_{2}\end{matrix}\right)=\left(\begin{matrix}2\left(-10\right)-\frac{1}{2}\times 20\\-\left(-10\right)+\frac{1}{2}\times 20\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}c_{1}\\c_{2}\end{matrix}\right)=\left(\begin{matrix}-30\\20\end{matrix}\right)$$

Extract the matrix elements c_{1} and c_{2}.

$$c_{1}=-30,c_{2}=20$$

Steps Using Elimination

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

$$c_{1}+c_{2}=-10$$

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

$$2c_{1}+4c_{2}=20$$

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.

$$c_{1}+c_{2}=-10,2c_{1}+4c_{2}=20$$

To make c_{1} and 2c_{1} equal, multiply all terms on each side of the first equation by 2 and all terms on each side of the second by 1.

$$2c_{1}+2c_{2}=2\left(-10\right),2c_{1}+4c_{2}=20$$

Simplify.

$$2c_{1}+2c_{2}=-20,2c_{1}+4c_{2}=20$$

Subtract 2c_{1}+4c_{2}=20 from 2c_{1}+2c_{2}=-20 by subtracting like terms on each side of the equal sign.

$$2c_{1}-2c_{1}+2c_{2}-4c_{2}=-20-20$$

Add 2c_{1} to -2c_{1}. Terms 2c_{1} and -2c_{1} cancel out, leaving an equation with only one variable that can be solved.

$$2c_{2}-4c_{2}=-20-20$$

Add 2c_{2} to -4c_{2}.

$$-2c_{2}=-20-20$$

Add -20 to -20.

$$-2c_{2}=-40$$

Divide both sides by -2.

$$c_{2}=20$$

Substitute 20 for c_{2} in 2c_{1}+4c_{2}=20. Because the resulting equation contains only one variable, you can solve for c_{1} directly.

$$2c_{1}+4\times 20=20$$

Multiply 4 times 20.

$$2c_{1}+80=20$$

Subtract 80 from both sides of the equation.

$$2c_{1}=-60$$

Divide both sides by 2.

$$c_{1}=-30$$

The system is now solved.

$$c_{1}=-30,c_{2}=20$$