Solve {l}{text{a)}8x-2y+10-3x-4y+6-3x-2y-2=3}{8x-3x-2y-4-2y+10+6-2=3}

$$\left. \begin{array} { l } { \text { a) } 8 x - 2 y + 10 - 3 x - 4 y + 6 - 3 x - 2 y - 2 = 3 } \\ { 8 x - 3 x - 2 y - 4 - 2 y + 10 + 6 - 2 = 3 } \end{array} \right.$$

Solve for x, y

$$x=-\frac{3}{8}=-0.375$$

$$y = \frac{41}{32} = 1\frac{9}{32} = 1.28125$$

Steps Using Substitution

Consider the first equation. Combine 8x and -3x to get 5x.

$$5x-2y+10-4y+6-3x-2y-2=3$$

Combine -2y and -4y to get -6y.

$$5x-6y+10+6-3x-2y-2=3$$

Add 10 and 6 to get 16.

$$5x-6y+16-3x-2y-2=3$$

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

$$2x-6y+16-2y-2=3$$

Combine -6y and -2y to get -8y.

$$2x-8y+16-2=3$$

Subtract 2 from 16 to get 14.

$$2x-8y+14=3$$

Subtract 14 from both sides.

$$2x-8y=3-14$$

Subtract 14 from 3 to get -11.

$$2x-8y=-11$$

Consider the second equation. Combine 8x and -3x to get 5x.

$$5x-2y-4-2y+10+6-2=3$$

Combine -2y and -2y to get -4y.

$$5x-4y-4+10+6-2=3$$

Add -4 and 10 to get 6.

$$5x-4y+6+6-2=3$$

Add 6 and 6 to get 12.

$$5x-4y+12-2=3$$

Subtract 2 from 12 to get 10.

$$5x-4y+10=3$$

Subtract 10 from both sides.

$$5x-4y=3-10$$

Subtract 10 from 3 to get -7.

$$5x-4y=-7$$

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.

$$2x-8y=-11,5x-4y=-7$$

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

$$2x-8y=-11$$

Add 8y to both sides of the equation.

$$2x=8y-11$$

Divide both sides by 2.

$$x=\frac{1}{2}\left(8y-11\right)$$

Multiply \frac{1}{2} times 8y-11.

$$x=4y-\frac{11}{2}$$

Substitute 4y-\frac{11}{2} for x in the other equation, 5x-4y=-7.

$$5\left(4y-\frac{11}{2}\right)-4y=-7$$

Multiply 5 times 4y-\frac{11}{2}.

$$20y-\frac{55}{2}-4y=-7$$

Add 20y to -4y.

$$16y-\frac{55}{2}=-7$$

Add \frac{55}{2} to both sides of the equation.

$$16y=\frac{41}{2}$$

Divide both sides by 16.

$$y=\frac{41}{32}$$

Substitute \frac{41}{32} for y in x=4y-\frac{11}{2}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=4\times \frac{41}{32}-\frac{11}{2}$$

Multiply 4 times \frac{41}{32}.

$$x=\frac{41}{8}-\frac{11}{2}$$

Add -\frac{11}{2} to \frac{41}{8} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{3}{8}$$

The system is now solved.

$$x=-\frac{3}{8},y=\frac{41}{32}$$

Steps Using Matrices

Consider the first equation. Combine 8x and -3x to get 5x.

$$5x-2y+10-4y+6-3x-2y-2=3$$

Combine -2y and -4y to get -6y.

$$5x-6y+10+6-3x-2y-2=3$$

Add 10 and 6 to get 16.

$$5x-6y+16-3x-2y-2=3$$

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

$$2x-6y+16-2y-2=3$$

Combine -6y and -2y to get -8y.

$$2x-8y+16-2=3$$

Subtract 2 from 16 to get 14.

$$2x-8y+14=3$$

Subtract 14 from both sides.

$$2x-8y=3-14$$

Subtract 14 from 3 to get -11.

$$2x-8y=-11$$

Consider the second equation. Combine 8x and -3x to get 5x.

$$5x-2y-4-2y+10+6-2=3$$

Combine -2y and -2y to get -4y.

$$5x-4y-4+10+6-2=3$$

Add -4 and 10 to get 6.

$$5x-4y+6+6-2=3$$

Add 6 and 6 to get 12.

$$5x-4y+12-2=3$$

Subtract 2 from 12 to get 10.

$$5x-4y+10=3$$

Subtract 10 from both sides.

$$5x-4y=3-10$$

Subtract 10 from 3 to get -7.

$$5x-4y=-7$$

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

$$2x-8y=-11,5x-4y=-7$$

Write the equations in matrix form.

$$\left(\begin{matrix}2&-8\\5&-4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-11\\-7\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&-8\\5&-4\end{matrix}\right).

$$inverse(\left(\begin{matrix}2&-8\\5&-4\end{matrix}\right))\left(\begin{matrix}2&-8\\5&-4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-8\\5&-4\end{matrix}\right))\left(\begin{matrix}-11\\-7\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-8\\5&-4\end{matrix}\right))\left(\begin{matrix}-11\\-7\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-8\\5&-4\end{matrix}\right))\left(\begin{matrix}-11\\-7\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{4}{2\left(-4\right)-\left(-8\times 5\right)}&-\frac{-8}{2\left(-4\right)-\left(-8\times 5\right)}\\-\frac{5}{2\left(-4\right)-\left(-8\times 5\right)}&\frac{2}{2\left(-4\right)-\left(-8\times 5\right)}\end{matrix}\right)\left(\begin{matrix}-11\\-7\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{8}&\frac{1}{4}\\-\frac{5}{32}&\frac{1}{16}\end{matrix}\right)\left(\begin{matrix}-11\\-7\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{8}\left(-11\right)+\frac{1}{4}\left(-7\right)\\-\frac{5}{32}\left(-11\right)+\frac{1}{16}\left(-7\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{3}{8}\\\frac{41}{32}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=-\frac{3}{8},y=\frac{41}{32}$$

Steps Using Elimination

Consider the first equation. Combine 8x and -3x to get 5x.

$$5x-2y+10-4y+6-3x-2y-2=3$$

Combine -2y and -4y to get -6y.

$$5x-6y+10+6-3x-2y-2=3$$

Add 10 and 6 to get 16.

$$5x-6y+16-3x-2y-2=3$$

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

$$2x-6y+16-2y-2=3$$

Combine -6y and -2y to get -8y.

$$2x-8y+16-2=3$$

Subtract 2 from 16 to get 14.

$$2x-8y+14=3$$

Subtract 14 from both sides.

$$2x-8y=3-14$$

Subtract 14 from 3 to get -11.

$$2x-8y=-11$$

Consider the second equation. Combine 8x and -3x to get 5x.

$$5x-2y-4-2y+10+6-2=3$$

Combine -2y and -2y to get -4y.

$$5x-4y-4+10+6-2=3$$

Add -4 and 10 to get 6.

$$5x-4y+6+6-2=3$$

Add 6 and 6 to get 12.

$$5x-4y+12-2=3$$

Subtract 2 from 12 to get 10.

$$5x-4y+10=3$$

Subtract 10 from both sides.

$$5x-4y=3-10$$

Subtract 10 from 3 to get -7.

$$5x-4y=-7$$

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.

$$2x-8y=-11,5x-4y=-7$$

To make 2x and 5x equal, multiply all terms on each side of the first equation by 5 and all terms on each side of the second by 2.

$$5\times 2x+5\left(-8\right)y=5\left(-11\right),2\times 5x+2\left(-4\right)y=2\left(-7\right)$$

Simplify.

$$10x-40y=-55,10x-8y=-14$$

Subtract 10x-8y=-14 from 10x-40y=-55 by subtracting like terms on each side of the equal sign.

$$10x-10x-40y+8y=-55+14$$

Add 10x to -10x. Terms 10x and -10x cancel out, leaving an equation with only one variable that can be solved.

$$-40y+8y=-55+14$$

Add -40y to 8y.

$$-32y=-55+14$$

Add -55 to 14.

$$-32y=-41$$

Divide both sides by -32.

$$y=\frac{41}{32}$$

Substitute \frac{41}{32} for y in 5x-4y=-7. Because the resulting equation contains only one variable, you can solve for x directly.

$$5x-4\times \frac{41}{32}=-7$$

Multiply -4 times \frac{41}{32}.

$$5x-\frac{41}{8}=-7$$

Add \frac{41}{8} to both sides of the equation.

$$5x=-\frac{15}{8}$$

Divide both sides by 5.

$$x=-\frac{3}{8}$$

The system is now solved.

$$x=-\frac{3}{8},y=\frac{41}{32}$$