Solve {l}{C_{1}+C_{2}=1}{-e/2+3/4+C_{2}e-C_{2}+frac{C_{1}e^2}{2}-1/2C_{1}=1/4e^2}

$$\left. \begin{array} { l } { C _ { 1 } + C _ { 2 } = 1 } \\ { - \frac { e } { 2 } + \frac { 3 } { 4 } + C _ { 2 } e - C _ { 2 } + \frac { C _ { 1 } e ^ { 2 } } { 2 } - \frac { 1 } { 2 } C _ { 1 } = \frac { 1 } { 4 } e ^ { 2 } } \end{array} \right.$$

Solve for C_1, C_2

$$C_{1}=\frac{1}{2}=0.5$$

$$C_{2}=\frac{1}{2}=0.5$$

Steps Using Substitution

Consider the second equation. Multiply both sides of the equation by 4, the least common multiple of 2,4.

$$-2e+3+4C_{2}e-4C_{2}+2C_{1}e^{2}-2C_{1}=1e^{2}$$

Add 2e to both sides.

$$3+4C_{2}e-4C_{2}+2C_{1}e^{2}-2C_{1}=1e^{2}+2e$$

Subtract 3 from both sides.

$$4C_{2}e-4C_{2}+2C_{1}e^{2}-2C_{1}=1e^{2}+2e-3$$

Reorder the terms.

$$2e^{2}C_{1}-2C_{1}+4eC_{2}-4C_{2}=e^{2}+2e-3$$

Combine all terms containing C_{1},C_{2}.

$$\left(2e^{2}-2\right)C_{1}+\left(4e-4\right)C_{2}=e^{2}+2e-3$$

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.

$$C_{1}+C_{2}=1,\left(2e^{2}-2\right)C_{1}+\left(4e-4\right)C_{2}=e^{2}+2e-3$$

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

$$C_{1}+C_{2}=1$$

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

$$C_{1}=-C_{2}+1$$

Substitute -C_{2}+1 for C_{1} in the other equation, \left(2e^{2}-2\right)C_{1}+\left(4e-4\right)C_{2}=e^{2}+2e-3.

$$\left(2e^{2}-2\right)\left(-C_{2}+1\right)+\left(4e-4\right)C_{2}=e^{2}+2e-3$$

Multiply 2e^{2}-2 times -C_{2}+1.

$$\left(-2e^{2}+2\right)C_{2}+2e^{2}-2+\left(4e-4\right)C_{2}=e^{2}+2e-3$$

Add -2C_{2}e^{2}+2C_{2} to 4\left(e-1\right)C_{2}.

$$\left(-2\left(e-1\right)^{2}\right)C_{2}+2e^{2}-2=e^{2}+2e-3$$

Subtract 2e^{2}-2 from both sides of the equation.

$$\left(-2\left(e-1\right)^{2}\right)C_{2}=-e^{2}+2e-1$$

Divide both sides by -2\left(e-1\right)^{2}.

$$C_{2}=\frac{1}{2}$$

Substitute \frac{1}{2} for C_{2} in C_{1}=-C_{2}+1. Because the resulting equation contains only one variable, you can solve for C_{1} directly.

$$C_{1}=-\frac{1}{2}+1$$

Add 1 to -\frac{1}{2}.

$$C_{1}=\frac{1}{2}$$

The system is now solved.

$$C_{1}=\frac{1}{2},C_{2}=\frac{1}{2}$$

Steps Using Matrices

Consider the second equation. Multiply both sides of the equation by 4, the least common multiple of 2,4.

$$-2e+3+4C_{2}e-4C_{2}+2C_{1}e^{2}-2C_{1}=1e^{2}$$

Add 2e to both sides.

$$3+4C_{2}e-4C_{2}+2C_{1}e^{2}-2C_{1}=1e^{2}+2e$$

Subtract 3 from both sides.

$$4C_{2}e-4C_{2}+2C_{1}e^{2}-2C_{1}=1e^{2}+2e-3$$

Reorder the terms.

$$2e^{2}C_{1}-2C_{1}+4eC_{2}-4C_{2}=e^{2}+2e-3$$

Combine all terms containing C_{1},C_{2}.

$$\left(2e^{2}-2\right)C_{1}+\left(4e-4\right)C_{2}=e^{2}+2e-3$$

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

$$C_{1}+C_{2}=1,\left(2e^{2}-2\right)C_{1}+\left(4e-4\right)C_{2}=e^{2}+2e-3$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&1\\2e^{2}-2&4e-4\end{matrix}\right)\left(\begin{matrix}C_{1}\\C_{2}\end{matrix}\right)=\left(\begin{matrix}1\\e^{2}+2e-3\end{matrix}\right)$$

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

$$inverse(\left(\begin{matrix}1&1\\2e^{2}-2&4e-4\end{matrix}\right))\left(\begin{matrix}1&1\\2e^{2}-2&4e-4\end{matrix}\right)\left(\begin{matrix}C_{1}\\C_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\2e^{2}-2&4e-4\end{matrix}\right))\left(\begin{matrix}1\\e^{2}+2e-3\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}C_{1}\\C_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\2e^{2}-2&4e-4\end{matrix}\right))\left(\begin{matrix}1\\e^{2}+2e-3\end{matrix}\right)$$

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

$$\left(\begin{matrix}C_{1}\\C_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\2e^{2}-2&4e-4\end{matrix}\right))\left(\begin{matrix}1\\e^{2}+2e-3\end{matrix}\right)$$

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.

$$\left(\begin{matrix}C_{1}\\C_{2}\end{matrix}\right)=\left(\begin{matrix}\frac{4e-4}{4e-4-\left(2e^{2}-2\right)}&-\frac{1}{4e-4-\left(2e^{2}-2\right)}\\-\frac{2e^{2}-2}{4e-4-\left(2e^{2}-2\right)}&\frac{1}{4e-4-\left(2e^{2}-2\right)}\end{matrix}\right)\left(\begin{matrix}1\\e^{2}+2e-3\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}C_{1}\\C_{2}\end{matrix}\right)=\left(\begin{matrix}-\frac{2}{e-1}&\frac{1}{2\left(e-1\right)^{2}}\\\frac{e+1}{e-1}&-\frac{1}{2\left(e-1\right)^{2}}\end{matrix}\right)\left(\begin{matrix}1\\e^{2}+2e-3\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}C_{1}\\C_{2}\end{matrix}\right)=\left(\begin{matrix}-\frac{2}{e-1}+\frac{1}{2\left(e-1\right)^{2}}\left(e^{2}+2e-3\right)\\\frac{e+1}{e-1}+\left(-\frac{1}{2\left(e-1\right)^{2}}\right)\left(e^{2}+2e-3\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}C_{1}\\C_{2}\end{matrix}\right)=\left(\begin{matrix}\frac{1}{2}\\\frac{1}{2}\end{matrix}\right)$$

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

$$C_{1}=\frac{1}{2},C_{2}=\frac{1}{2}$$

Steps Using Elimination

Consider the second equation. Multiply both sides of the equation by 4, the least common multiple of 2,4.

$$-2e+3+4C_{2}e-4C_{2}+2C_{1}e^{2}-2C_{1}=1e^{2}$$

Add 2e to both sides.

$$3+4C_{2}e-4C_{2}+2C_{1}e^{2}-2C_{1}=1e^{2}+2e$$

Subtract 3 from both sides.

$$4C_{2}e-4C_{2}+2C_{1}e^{2}-2C_{1}=1e^{2}+2e-3$$

Reorder the terms.

$$2e^{2}C_{1}-2C_{1}+4eC_{2}-4C_{2}=e^{2}+2e-3$$

Combine all terms containing C_{1},C_{2}.

$$\left(2e^{2}-2\right)C_{1}+\left(4e-4\right)C_{2}=e^{2}+2e-3$$

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.

$$C_{1}+C_{2}=1,\left(2e^{2}-2\right)C_{1}+\left(4e-4\right)C_{2}=e^{2}+2e-3$$

To make C_{1} and 2C_{1}\left(e^{2}-1\right) equal, multiply all terms on each side of the first equation by 2e^{2}-2 and all terms on each side of the second by 1.

$$\left(2e^{2}-2\right)C_{1}+\left(2e^{2}-2\right)C_{2}=2e^{2}-2,\left(2e^{2}-2\right)C_{1}+\left(4e-4\right)C_{2}=e^{2}+2e-3$$

Subtract \left(2e^{2}-2\right)C_{1}+\left(4e-4\right)C_{2}=e^{2}+2e-3 from \left(2e^{2}-2\right)C_{1}+\left(2e^{2}-2\right)C_{2}=2e^{2}-2 by subtracting like terms on each side of the equal sign.

$$\left(2e^{2}-2\right)C_{1}+\left(-2e^{2}+2\right)C_{1}+\left(2e^{2}-2\right)C_{2}+\left(4-4e\right)C_{2}=2e^{2}-2-e^{2}+3-2e$$

Add 2C_{1}\left(e^{2}-1\right) to -2C_{1}e^{2}+2C_{1}. Terms 2C_{1}\left(e^{2}-1\right) and -2C_{1}e^{2}+2C_{1} cancel out, leaving an equation with only one variable that can be solved.

$$\left(2e^{2}-2\right)C_{2}+\left(4-4e\right)C_{2}=2e^{2}-2-e^{2}+3-2e$$

Add 2C_{2}\left(e^{2}-1\right) to -4C_{2}e+4C_{2}.

$$2\left(e-1\right)^{2}C_{2}=2e^{2}-2-e^{2}+3-2e$$

Add 2e^{2}-2 to -e^{2}-2e+3.

$$2\left(e-1\right)^{2}C_{2}=\left(e-1\right)^{2}$$

Divide both sides by 2\left(e-1\right)^{2}.

$$C_{2}=\frac{1}{2}$$

Substitute \frac{1}{2} for C_{2} in \left(2e^{2}-2\right)C_{1}+\left(4e-4\right)C_{2}=e^{2}+2e-3. Because the resulting equation contains only one variable, you can solve for C_{1} directly.

$$\left(2e^{2}-2\right)C_{1}+\left(4e-4\right)\times \frac{1}{2}=e^{2}+2e-3$$

Multiply 4e-4 times \frac{1}{2}.

$$\left(2e^{2}-2\right)C_{1}+2e-2=e^{2}+2e-3$$

Subtract -2+2e from both sides of the equation.

$$\left(2e^{2}-2\right)C_{1}=e^{2}-1$$

Divide both sides by 2e^{2}-2.

$$C_{1}=\frac{1}{2}$$

The system is now solved.

$$C_{1}=\frac{1}{2},C_{2}=\frac{1}{2}$$