Solve {c}{3y-9=4x-4y+5}{5y-3=-(3x+2)+1+2y}

$$\left\{ \begin{array} { c } { 3 y - 9 = 4 x - 4 y + 5 } \\ { 5 y - 3 = - ( 3 x + 2 ) + 1 + 2 y } \end{array} \right.$$

Solve for y, x

$$x=-\frac{28}{33}\approx -0.848484848$$

$$y = \frac{50}{33} = 1\frac{17}{33} \approx 1.515151515$$

Steps Using Substitution

Consider the first equation. Subtract 4x from both sides.

$$3y-9-4x=-4y+5$$

Add 4y to both sides.

$$3y-9-4x+4y=5$$

Combine 3y and 4y to get 7y.

$$7y-9-4x=5$$

Add 9 to both sides.

$$7y-4x=5+9$$

Add 5 and 9 to get 14.

$$7y-4x=14$$

Consider the second equation. To find the opposite of 3x+2, find the opposite of each term.

$$5y-3=-3x-2+1+2y$$

Add -2 and 1 to get -1.

$$5y-3=-3x-1+2y$$

Add 3x to both sides.

$$5y-3+3x=-1+2y$$

Subtract 2y from both sides.

$$5y-3+3x-2y=-1$$

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

$$3y-3+3x=-1$$

Add 3 to both sides.

$$3y+3x=-1+3$$

Add -1 and 3 to get 2.

$$3y+3x=2$$

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.

$$7y-4x=14,3y+3x=2$$

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

$$7y-4x=14$$

Add 4x to both sides of the equation.

$$7y=4x+14$$

Divide both sides by 7.

$$y=\frac{1}{7}\left(4x+14\right)$$

Multiply \frac{1}{7} times 4x+14.

$$y=\frac{4}{7}x+2$$

Substitute \frac{4x}{7}+2 for y in the other equation, 3y+3x=2.

$$3\left(\frac{4}{7}x+2\right)+3x=2$$

Multiply 3 times \frac{4x}{7}+2.

$$\frac{12}{7}x+6+3x=2$$

Add \frac{12x}{7} to 3x.

$$\frac{33}{7}x+6=2$$

Subtract 6 from both sides of the equation.

$$\frac{33}{7}x=-4$$

Divide both sides of the equation by \frac{33}{7}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$x=-\frac{28}{33}$$

Substitute -\frac{28}{33} for x in y=\frac{4}{7}x+2. Because the resulting equation contains only one variable, you can solve for y directly.

$$y=\frac{4}{7}\left(-\frac{28}{33}\right)+2$$

Multiply \frac{4}{7} times -\frac{28}{33} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$y=-\frac{16}{33}+2$$

Add 2 to -\frac{16}{33}.

$$y=\frac{50}{33}$$

The system is now solved.

$$y=\frac{50}{33},x=-\frac{28}{33}$$

Steps Using Matrices

Consider the first equation. Subtract 4x from both sides.

$$3y-9-4x=-4y+5$$

Add 4y to both sides.

$$3y-9-4x+4y=5$$

Combine 3y and 4y to get 7y.

$$7y-9-4x=5$$

Add 9 to both sides.

$$7y-4x=5+9$$

Add 5 and 9 to get 14.

$$7y-4x=14$$

Consider the second equation. To find the opposite of 3x+2, find the opposite of each term.

$$5y-3=-3x-2+1+2y$$

Add -2 and 1 to get -1.

$$5y-3=-3x-1+2y$$

Add 3x to both sides.

$$5y-3+3x=-1+2y$$

Subtract 2y from both sides.

$$5y-3+3x-2y=-1$$

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

$$3y-3+3x=-1$$

Add 3 to both sides.

$$3y+3x=-1+3$$

Add -1 and 3 to get 2.

$$3y+3x=2$$

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

$$7y-4x=14,3y+3x=2$$

Write the equations in matrix form.

$$\left(\begin{matrix}7&-4\\3&3\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}14\\2\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}7&-4\\3&3\end{matrix}\right).

$$inverse(\left(\begin{matrix}7&-4\\3&3\end{matrix}\right))\left(\begin{matrix}7&-4\\3&3\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}7&-4\\3&3\end{matrix}\right))\left(\begin{matrix}14\\2\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}7&-4\\3&3\end{matrix}\right))\left(\begin{matrix}14\\2\end{matrix}\right)$$

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

$$\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}7&-4\\3&3\end{matrix}\right))\left(\begin{matrix}14\\2\end{matrix}\right)$$

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.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{3}{7\times 3-\left(-4\times 3\right)}&-\frac{-4}{7\times 3-\left(-4\times 3\right)}\\-\frac{3}{7\times 3-\left(-4\times 3\right)}&\frac{7}{7\times 3-\left(-4\times 3\right)}\end{matrix}\right)\left(\begin{matrix}14\\2\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{1}{11}&\frac{4}{33}\\-\frac{1}{11}&\frac{7}{33}\end{matrix}\right)\left(\begin{matrix}14\\2\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{1}{11}\times 14+\frac{4}{33}\times 2\\-\frac{1}{11}\times 14+\frac{7}{33}\times 2\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{50}{33}\\-\frac{28}{33}\end{matrix}\right)$$

Extract the matrix elements y and x.

$$y=\frac{50}{33},x=-\frac{28}{33}$$

Steps Using Elimination

Consider the first equation. Subtract 4x from both sides.

$$3y-9-4x=-4y+5$$

Add 4y to both sides.

$$3y-9-4x+4y=5$$

Combine 3y and 4y to get 7y.

$$7y-9-4x=5$$

Add 9 to both sides.

$$7y-4x=5+9$$

Add 5 and 9 to get 14.

$$7y-4x=14$$

Consider the second equation. To find the opposite of 3x+2, find the opposite of each term.

$$5y-3=-3x-2+1+2y$$

Add -2 and 1 to get -1.

$$5y-3=-3x-1+2y$$

Add 3x to both sides.

$$5y-3+3x=-1+2y$$

Subtract 2y from both sides.

$$5y-3+3x-2y=-1$$

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

$$3y-3+3x=-1$$

Add 3 to both sides.

$$3y+3x=-1+3$$

Add -1 and 3 to get 2.

$$3y+3x=2$$

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.

$$7y-4x=14,3y+3x=2$$

To make 7y 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 7.

$$3\times 7y+3\left(-4\right)x=3\times 14,7\times 3y+7\times 3x=7\times 2$$

Simplify.

$$21y-12x=42,21y+21x=14$$

Subtract 21y+21x=14 from 21y-12x=42 by subtracting like terms on each side of the equal sign.

$$21y-21y-12x-21x=42-14$$

Add 21y to -21y. Terms 21y and -21y cancel out, leaving an equation with only one variable that can be solved.

$$-12x-21x=42-14$$

Add -12x to -21x.

$$-33x=42-14$$

Add 42 to -14.

$$-33x=28$$

Divide both sides by -33.

$$x=-\frac{28}{33}$$

Substitute -\frac{28}{33} for x in 3y+3x=2. Because the resulting equation contains only one variable, you can solve for y directly.

$$3y+3\left(-\frac{28}{33}\right)=2$$

Multiply 3 times -\frac{28}{33}.

$$3y-\frac{28}{11}=2$$

Add \frac{28}{11} to both sides of the equation.

$$3y=\frac{50}{11}$$

Divide both sides by 3.

$$y=\frac{50}{33}$$

The system is now solved.

$$y=\frac{50}{33},x=-\frac{28}{33}$$