Solve {c}{27/5+4/5*y=-4*x+3/5}{8*x=-8-3*y}

$$\left.\begin{array} { c } { \frac { 27 } { 5 } + \frac { 4 } { 5 } \cdot y = \frac { - 4 \cdot x + 3 } { 5 } } \\ { 8 \cdot x = - 8 - 3 \cdot y } \end{array} \right\}$$

Solve for y, x

$$x=2$$

$$y=-8$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 5.

$$27+4y=-4x+3$$

Add 4x to both sides.

$$27+4y+4x=3$$

Subtract 27 from both sides.

$$4y+4x=3-27$$

Subtract 27 from 3 to get -24.

$$4y+4x=-24$$

Consider the second equation. Add 3y to both sides.

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

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.

$$4y+4x=-24,3y+8x=-8$$

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

$$4y+4x=-24$$

Subtract 4x from both sides of the equation.

$$4y=-4x-24$$

Divide both sides by 4.

$$y=\frac{1}{4}\left(-4x-24\right)$$

Multiply \frac{1}{4} times -4x-24.

$$y=-x-6$$

Substitute -x-6 for y in the other equation, 3y+8x=-8.

$$3\left(-x-6\right)+8x=-8$$

Multiply 3 times -x-6.

$$-3x-18+8x=-8$$

Add -3x to 8x.

$$5x-18=-8$$

Add 18 to both sides of the equation.

$$5x=10$$

Divide both sides by 5.

$$x=2$$

Substitute 2 for x in y=-x-6. Because the resulting equation contains only one variable, you can solve for y directly.

$$y=-2-6$$

Add -6 to -2.

$$y=-8$$

The system is now solved.

$$y=-8,x=2$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 5.

$$27+4y=-4x+3$$

Add 4x to both sides.

$$27+4y+4x=3$$

Subtract 27 from both sides.

$$4y+4x=3-27$$

Subtract 27 from 3 to get -24.

$$4y+4x=-24$$

Consider the second equation. Add 3y to both sides.

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

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

$$4y+4x=-24,3y+8x=-8$$

Write the equations in matrix form.

$$\left(\begin{matrix}4&4\\3&8\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-24\\-8\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}4&4\\3&8\end{matrix}\right).

$$inverse(\left(\begin{matrix}4&4\\3&8\end{matrix}\right))\left(\begin{matrix}4&4\\3&8\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}4&4\\3&8\end{matrix}\right))\left(\begin{matrix}-24\\-8\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}4&4\\3&8\end{matrix}\right))\left(\begin{matrix}-24\\-8\end{matrix}\right)$$

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

$$\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}4&4\\3&8\end{matrix}\right))\left(\begin{matrix}-24\\-8\end{matrix}\right)$$

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.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{8}{4\times 8-4\times 3}&-\frac{4}{4\times 8-4\times 3}\\-\frac{3}{4\times 8-4\times 3}&\frac{4}{4\times 8-4\times 3}\end{matrix}\right)\left(\begin{matrix}-24\\-8\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{2}{5}&-\frac{1}{5}\\-\frac{3}{20}&\frac{1}{5}\end{matrix}\right)\left(\begin{matrix}-24\\-8\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{2}{5}\left(-24\right)-\frac{1}{5}\left(-8\right)\\-\frac{3}{20}\left(-24\right)+\frac{1}{5}\left(-8\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-8\\2\end{matrix}\right)$$

Extract the matrix elements y and x.

$$y=-8,x=2$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 5.

$$27+4y=-4x+3$$

Add 4x to both sides.

$$27+4y+4x=3$$

Subtract 27 from both sides.

$$4y+4x=3-27$$

Subtract 27 from 3 to get -24.

$$4y+4x=-24$$

Consider the second equation. Add 3y to both sides.

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

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.

$$4y+4x=-24,3y+8x=-8$$

To make 4y and 3y equal, multiply all terms on each side of the first equation by 3 and all terms on each side of the second by 4.

$$3\times 4y+3\times 4x=3\left(-24\right),4\times 3y+4\times 8x=4\left(-8\right)$$

Simplify.

$$12y+12x=-72,12y+32x=-32$$

Subtract 12y+32x=-32 from 12y+12x=-72 by subtracting like terms on each side of the equal sign.

$$12y-12y+12x-32x=-72+32$$

Add 12y to -12y. Terms 12y and -12y cancel out, leaving an equation with only one variable that can be solved.

$$12x-32x=-72+32$$

Add 12x to -32x.

$$-20x=-72+32$$

Add -72 to 32.

$$-20x=-40$$

Divide both sides by -20.

$$x=2$$

Substitute 2 for x in 3y+8x=-8. Because the resulting equation contains only one variable, you can solve for y directly.

$$3y+8\times 2=-8$$

Multiply 8 times 2.

$$3y+16=-8$$

Subtract 16 from both sides of the equation.

$$3y=-24$$

Divide both sides by 3.

$$y=-8$$

The system is now solved.

$$y=-8,x=2$$