Solve {l}{6x+5y+80=1400}{4x+2y+50=1150}

$$\left\{ \begin{array} { l } { 6 x + 5 y + 80 = 1400 } \\ { 4 x + 2 y + 50 = 1150 } \end{array} \right.$$

Solve for x, y

$$x = \frac{715}{2} = 357\frac{1}{2} = 357.5$$

$$y=-165$$

Steps Using Substitution

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.

$$6x+5y+80=1400,4x+2y+50=1150$$

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

$$6x+5y+80=1400$$

Subtract 80 from both sides of the equation.

$$6x+5y=1320$$

Subtract 5y from both sides of the equation.

$$6x=-5y+1320$$

Divide both sides by 6.

$$x=\frac{1}{6}\left(-5y+1320\right)$$

Multiply \frac{1}{6} times -5y+1320.

$$x=-\frac{5}{6}y+220$$

Substitute -\frac{5y}{6}+220 for x in the other equation, 4x+2y+50=1150.

$$4\left(-\frac{5}{6}y+220\right)+2y+50=1150$$

Multiply 4 times -\frac{5y}{6}+220.

$$-\frac{10}{3}y+880+2y+50=1150$$

Add -\frac{10y}{3} to 2y.

$$-\frac{4}{3}y+880+50=1150$$

Add 880 to 50.

$$-\frac{4}{3}y+930=1150$$

Subtract 930 from both sides of the equation.

$$-\frac{4}{3}y=220$$

Divide both sides of the equation by -\frac{4}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=-165$$

Substitute -165 for y in x=-\frac{5}{6}y+220. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{5}{6}\left(-165\right)+220$$

Multiply -\frac{5}{6} times -165.

$$x=\frac{275}{2}+220$$

Add 220 to \frac{275}{2}.

$$x=\frac{715}{2}$$

The system is now solved.

$$x=\frac{715}{2},y=-165$$

Steps Using Matrices

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

$$6x+5y+80=1400,4x+2y+50=1150$$

Write the equations in matrix form.

$$\left(\begin{matrix}6&5\\4&2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}1320\\1100\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}6&5\\4&2\end{matrix}\right).

$$inverse(\left(\begin{matrix}6&5\\4&2\end{matrix}\right))\left(\begin{matrix}6&5\\4&2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}6&5\\4&2\end{matrix}\right))\left(\begin{matrix}1320\\1100\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}6&5\\4&2\end{matrix}\right))\left(\begin{matrix}1320\\1100\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}6&5\\4&2\end{matrix}\right))\left(\begin{matrix}1320\\1100\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{6\times 2-5\times 4}&-\frac{5}{6\times 2-5\times 4}\\-\frac{4}{6\times 2-5\times 4}&\frac{6}{6\times 2-5\times 4}\end{matrix}\right)\left(\begin{matrix}1320\\1100\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{4}&\frac{5}{8}\\\frac{1}{2}&-\frac{3}{4}\end{matrix}\right)\left(\begin{matrix}1320\\1100\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{4}\times 1320+\frac{5}{8}\times 1100\\\frac{1}{2}\times 1320-\frac{3}{4}\times 1100\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{715}{2}\\-165\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{715}{2},y=-165$$

Steps Using Elimination

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.

$$6x+5y+80=1400,4x+2y+50=1150$$

To make 6x and 4x equal, multiply all terms on each side of the first equation by 4 and all terms on each side of the second by 6.

$$4\times 6x+4\times 5y+4\times 80=4\times 1400,6\times 4x+6\times 2y+6\times 50=6\times 1150$$

Simplify.

$$24x+20y+320=5600,24x+12y+300=6900$$

Subtract 24x+12y+300=6900 from 24x+20y+320=5600 by subtracting like terms on each side of the equal sign.

$$24x-24x+20y-12y+320-300=5600-6900$$

Add 24x to -24x. Terms 24x and -24x cancel out, leaving an equation with only one variable that can be solved.

$$20y-12y+320-300=5600-6900$$

Add 20y to -12y.

$$8y+320-300=5600-6900$$

Add 320 to -300.

$$8y+20=5600-6900$$

Add 5600 to -6900.

$$8y+20=-1300$$

Subtract 20 from both sides of the equation.

$$8y=-1320$$

Divide both sides by 8.

$$y=-165$$

Substitute -165 for y in 4x+2y+50=1150. Because the resulting equation contains only one variable, you can solve for x directly.

$$4x+2\left(-165\right)+50=1150$$

Multiply 2 times -165.

$$4x-330+50=1150$$

Add -330 to 50.

$$4x-280=1150$$

Add 280 to both sides of the equation.

$$4x=1430$$

Divide both sides by 4.

$$x=\frac{715}{2}$$

The system is now solved.

$$x=\frac{715}{2},y=-165$$