Solve {l}{3x+2y=5x+2}{213x+2y=11x+7}

$$\left\{ \begin{array} { l } { 3 x + 2 y = 5 x + 2 } \\ { 213 x + 2 y = 11 x + 7 } \end{array} \right.$$

Solve for x, y

$$x=\frac{5}{204}\approx 0.024509804$$

$$y = \frac{209}{204} = 1\frac{5}{204} \approx 1.024509804$$

Steps Using Substitution

Consider the first equation. Subtract 5x from both sides.

$$3x+2y-5x=2$$

Combine 3x and -5x to get -2x.

$$-2x+2y=2$$

Consider the second equation. Subtract 11x from both sides.

$$213x+2y-11x=7$$

Combine 213x and -11x to get 202x.

$$202x+2y=7$$

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.

$$-2x+2y=2,202x+2y=7$$

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

$$-2x+2y=2$$

Subtract 2y from both sides of the equation.

$$-2x=-2y+2$$

Divide both sides by -2.

$$x=-\frac{1}{2}\left(-2y+2\right)$$

Multiply -\frac{1}{2} times -2y+2.

$$x=y-1$$

Substitute y-1 for x in the other equation, 202x+2y=7.

$$202\left(y-1\right)+2y=7$$

Multiply 202 times y-1.

$$202y-202+2y=7$$

Add 202y to 2y.

$$204y-202=7$$

Add 202 to both sides of the equation.

$$204y=209$$

Divide both sides by 204.

$$y=\frac{209}{204}$$

Substitute \frac{209}{204} for y in x=y-1. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{209}{204}-1$$

Add -1 to \frac{209}{204}.

$$x=\frac{5}{204}$$

The system is now solved.

$$x=\frac{5}{204},y=\frac{209}{204}$$

Steps Using Matrices

Consider the first equation. Subtract 5x from both sides.

$$3x+2y-5x=2$$

Combine 3x and -5x to get -2x.

$$-2x+2y=2$$

Consider the second equation. Subtract 11x from both sides.

$$213x+2y-11x=7$$

Combine 213x and -11x to get 202x.

$$202x+2y=7$$

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

$$-2x+2y=2,202x+2y=7$$

Write the equations in matrix form.

$$\left(\begin{matrix}-2&2\\202&2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}2\\7\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}-2&2\\202&2\end{matrix}\right).

$$inverse(\left(\begin{matrix}-2&2\\202&2\end{matrix}\right))\left(\begin{matrix}-2&2\\202&2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-2&2\\202&2\end{matrix}\right))\left(\begin{matrix}2\\7\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-2&2\\202&2\end{matrix}\right))\left(\begin{matrix}2\\7\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-2&2\\202&2\end{matrix}\right))\left(\begin{matrix}2\\7\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{-2\times 2-2\times 202}&-\frac{2}{-2\times 2-2\times 202}\\-\frac{202}{-2\times 2-2\times 202}&-\frac{2}{-2\times 2-2\times 202}\end{matrix}\right)\left(\begin{matrix}2\\7\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{204}&\frac{1}{204}\\\frac{101}{204}&\frac{1}{204}\end{matrix}\right)\left(\begin{matrix}2\\7\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{204}\times 2+\frac{1}{204}\times 7\\\frac{101}{204}\times 2+\frac{1}{204}\times 7\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{204}\\\frac{209}{204}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{5}{204},y=\frac{209}{204}$$

Steps Using Elimination

Consider the first equation. Subtract 5x from both sides.

$$3x+2y-5x=2$$

Combine 3x and -5x to get -2x.

$$-2x+2y=2$$

Consider the second equation. Subtract 11x from both sides.

$$213x+2y-11x=7$$

Combine 213x and -11x to get 202x.

$$202x+2y=7$$

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.

$$-2x+2y=2,202x+2y=7$$

Subtract 202x+2y=7 from -2x+2y=2 by subtracting like terms on each side of the equal sign.

$$-2x-202x+2y-2y=2-7$$

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

$$-2x-202x=2-7$$

Add -2x to -202x.

$$-204x=2-7$$

Add 2 to -7.

$$-204x=-5$$

Divide both sides by -204.

$$x=\frac{5}{204}$$

Substitute \frac{5}{204} for x in 202x+2y=7. Because the resulting equation contains only one variable, you can solve for y directly.

$$202\times \frac{5}{204}+2y=7$$

Multiply 202 times \frac{5}{204}.

$$\frac{505}{102}+2y=7$$

Subtract \frac{505}{102} from both sides of the equation.

$$2y=\frac{209}{102}$$

Divide both sides by 2.

$$y=\frac{209}{204}$$

The system is now solved.

$$x=\frac{5}{204},y=\frac{209}{204}$$