Solve {l}{2y+x+y=16}{200y+10x+y-297=100x+10x+2y}

$$\left\{ \begin{array} { l } { 2 y + x + y = 16 } \\ { 200 y + 10 x + y - 297 = 100 x + 10 x + 2 y } \end{array} \right.$$

Solve for y, x

$$x = \frac{2293}{499} = 4\frac{297}{499} \approx 4.595190381$$

$$y = \frac{1897}{499} = 3\frac{400}{499} \approx 3.801603206$$

Steps Using Substitution

Consider the first equation. Combine 2y and y to get 3y.

$$3y+x=16$$

Consider the second equation. Combine 200y and y to get 201y.

$$201y+10x-297=100x+10x+2y$$

Combine 100x and 10x to get 110x.

$$201y+10x-297=110x+2y$$

Subtract 110x from both sides.

$$201y+10x-297-110x=2y$$

Combine 10x and -110x to get -100x.

$$201y-100x-297=2y$$

Subtract 2y from both sides.

$$201y-100x-297-2y=0$$

Combine 201y and -2y to get 199y.

$$199y-100x-297=0$$

Add 297 to both sides. Anything plus zero gives itself.

$$199y-100x=297$$

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.

$$3y+x=16,199y-100x=297$$

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

$$3y+x=16$$

Subtract x from both sides of the equation.

$$3y=-x+16$$

Divide both sides by 3.

$$y=\frac{1}{3}\left(-x+16\right)$$

Multiply \frac{1}{3} times -x+16.

$$y=-\frac{1}{3}x+\frac{16}{3}$$

Substitute \frac{-x+16}{3} for y in the other equation, 199y-100x=297.

$$199\left(-\frac{1}{3}x+\frac{16}{3}\right)-100x=297$$

Multiply 199 times \frac{-x+16}{3}.

$$-\frac{199}{3}x+\frac{3184}{3}-100x=297$$

Add -\frac{199x}{3} to -100x.

$$-\frac{499}{3}x+\frac{3184}{3}=297$$

Subtract \frac{3184}{3} from both sides of the equation.

$$-\frac{499}{3}x=-\frac{2293}{3}$$

Divide both sides of the equation by -\frac{499}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$x=\frac{2293}{499}$$

Substitute \frac{2293}{499} for x in y=-\frac{1}{3}x+\frac{16}{3}. Because the resulting equation contains only one variable, you can solve for y directly.

$$y=-\frac{1}{3}\times \frac{2293}{499}+\frac{16}{3}$$

Multiply -\frac{1}{3} times \frac{2293}{499} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$y=-\frac{2293}{1497}+\frac{16}{3}$$

Add \frac{16}{3} to -\frac{2293}{1497} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{1897}{499}$$

The system is now solved.

$$y=\frac{1897}{499},x=\frac{2293}{499}$$

Steps Using Matrices

Consider the first equation. Combine 2y and y to get 3y.

$$3y+x=16$$

Consider the second equation. Combine 200y and y to get 201y.

$$201y+10x-297=100x+10x+2y$$

Combine 100x and 10x to get 110x.

$$201y+10x-297=110x+2y$$

Subtract 110x from both sides.

$$201y+10x-297-110x=2y$$

Combine 10x and -110x to get -100x.

$$201y-100x-297=2y$$

Subtract 2y from both sides.

$$201y-100x-297-2y=0$$

Combine 201y and -2y to get 199y.

$$199y-100x-297=0$$

Add 297 to both sides. Anything plus zero gives itself.

$$199y-100x=297$$

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

$$3y+x=16,199y-100x=297$$

Write the equations in matrix form.

$$\left(\begin{matrix}3&1\\199&-100\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}16\\297\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}3&1\\199&-100\end{matrix}\right).

$$inverse(\left(\begin{matrix}3&1\\199&-100\end{matrix}\right))\left(\begin{matrix}3&1\\199&-100\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}3&1\\199&-100\end{matrix}\right))\left(\begin{matrix}16\\297\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}3&1\\199&-100\end{matrix}\right))\left(\begin{matrix}16\\297\end{matrix}\right)$$

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

$$\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}3&1\\199&-100\end{matrix}\right))\left(\begin{matrix}16\\297\end{matrix}\right)$$

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.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-\frac{100}{3\left(-100\right)-199}&-\frac{1}{3\left(-100\right)-199}\\-\frac{199}{3\left(-100\right)-199}&\frac{3}{3\left(-100\right)-199}\end{matrix}\right)\left(\begin{matrix}16\\297\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{100}{499}&\frac{1}{499}\\\frac{199}{499}&-\frac{3}{499}\end{matrix}\right)\left(\begin{matrix}16\\297\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{100}{499}\times 16+\frac{1}{499}\times 297\\\frac{199}{499}\times 16-\frac{3}{499}\times 297\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{1897}{499}\\\frac{2293}{499}\end{matrix}\right)$$

Extract the matrix elements y and x.

$$y=\frac{1897}{499},x=\frac{2293}{499}$$

Steps Using Elimination

Consider the first equation. Combine 2y and y to get 3y.

$$3y+x=16$$

Consider the second equation. Combine 200y and y to get 201y.

$$201y+10x-297=100x+10x+2y$$

Combine 100x and 10x to get 110x.

$$201y+10x-297=110x+2y$$

Subtract 110x from both sides.

$$201y+10x-297-110x=2y$$

Combine 10x and -110x to get -100x.

$$201y-100x-297=2y$$

Subtract 2y from both sides.

$$201y-100x-297-2y=0$$

Combine 201y and -2y to get 199y.

$$199y-100x-297=0$$

Add 297 to both sides. Anything plus zero gives itself.

$$199y-100x=297$$

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.

$$3y+x=16,199y-100x=297$$

To make 3y and 199y equal, multiply all terms on each side of the first equation by 199 and all terms on each side of the second by 3.

$$199\times 3y+199x=199\times 16,3\times 199y+3\left(-100\right)x=3\times 297$$

Simplify.

$$597y+199x=3184,597y-300x=891$$

Subtract 597y-300x=891 from 597y+199x=3184 by subtracting like terms on each side of the equal sign.

$$597y-597y+199x+300x=3184-891$$

Add 597y to -597y. Terms 597y and -597y cancel out, leaving an equation with only one variable that can be solved.

$$199x+300x=3184-891$$

Add 199x to 300x.

$$499x=3184-891$$

Add 3184 to -891.

$$499x=2293$$

Divide both sides by 499.

$$x=\frac{2293}{499}$$

Substitute \frac{2293}{499} for x in 199y-100x=297. Because the resulting equation contains only one variable, you can solve for y directly.

$$199y-100\times \frac{2293}{499}=297$$

Multiply -100 times \frac{2293}{499}.

$$199y-\frac{229300}{499}=297$$

Add \frac{229300}{499} to both sides of the equation.

$$199y=\frac{377503}{499}$$

Divide both sides by 199.

$$y=\frac{1897}{499}$$

The system is now solved.

$$y=\frac{1897}{499},x=\frac{2293}{499}$$