Solve {l}{93y+4x=23}{3x+2y+43y=24}

$$\left. \begin{array} { l } { 93 y + 4 x = 23 } \\ { 3 x + 2 y + 43 y = 24 } \end{array} \right.$$

Solve for y, x

$$x = \frac{133}{11} = 12\frac{1}{11} \approx 12.090909091$$

$$y=-\frac{3}{11}\approx -0.272727273$$

Steps Using Substitution

Consider the second equation. Combine 2y and 43y to get 45y.

$$3x+45y=24$$

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.

$$93y+4x=23,45y+3x=24$$

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

$$93y+4x=23$$

Subtract 4x from both sides of the equation.

$$93y=-4x+23$$

Divide both sides by 93.

$$y=\frac{1}{93}\left(-4x+23\right)$$

Multiply \frac{1}{93} times -4x+23.

$$y=-\frac{4}{93}x+\frac{23}{93}$$

Substitute \frac{-4x+23}{93} for y in the other equation, 45y+3x=24.

$$45\left(-\frac{4}{93}x+\frac{23}{93}\right)+3x=24$$

Multiply 45 times \frac{-4x+23}{93}.

$$-\frac{60}{31}x+\frac{345}{31}+3x=24$$

Add -\frac{60x}{31} to 3x.

$$\frac{33}{31}x+\frac{345}{31}=24$$

Subtract \frac{345}{31} from both sides of the equation.

$$\frac{33}{31}x=\frac{399}{31}$$

Divide both sides of the equation by \frac{33}{31}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$x=\frac{133}{11}$$

Substitute \frac{133}{11} for x in y=-\frac{4}{93}x+\frac{23}{93}. Because the resulting equation contains only one variable, you can solve for y directly.

$$y=-\frac{4}{93}\times \frac{133}{11}+\frac{23}{93}$$

Multiply -\frac{4}{93} times \frac{133}{11} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$y=-\frac{532}{1023}+\frac{23}{93}$$

Add \frac{23}{93} to -\frac{532}{1023} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

The system is now solved.

$$y=-\frac{3}{11},x=\frac{133}{11}$$

Steps Using Matrices

Consider the second equation. Combine 2y and 43y to get 45y.

$$3x+45y=24$$

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

$$93y+4x=23,45y+3x=24$$

Write the equations in matrix form.

$$\left(\begin{matrix}93&4\\45&3\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}23\\24\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}93&4\\45&3\end{matrix}\right).

$$inverse(\left(\begin{matrix}93&4\\45&3\end{matrix}\right))\left(\begin{matrix}93&4\\45&3\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}93&4\\45&3\end{matrix}\right))\left(\begin{matrix}23\\24\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}93&4\\45&3\end{matrix}\right))\left(\begin{matrix}23\\24\end{matrix}\right)$$

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

$$\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}93&4\\45&3\end{matrix}\right))\left(\begin{matrix}23\\24\end{matrix}\right)$$

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.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{3}{93\times 3-4\times 45}&-\frac{4}{93\times 3-4\times 45}\\-\frac{45}{93\times 3-4\times 45}&\frac{93}{93\times 3-4\times 45}\end{matrix}\right)\left(\begin{matrix}23\\24\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{1}{33}&-\frac{4}{99}\\-\frac{5}{11}&\frac{31}{33}\end{matrix}\right)\left(\begin{matrix}23\\24\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{1}{33}\times 23-\frac{4}{99}\times 24\\-\frac{5}{11}\times 23+\frac{31}{33}\times 24\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-\frac{3}{11}\\\frac{133}{11}\end{matrix}\right)$$

Extract the matrix elements y and x.

$$y=-\frac{3}{11},x=\frac{133}{11}$$

Steps Using Elimination

Consider the second equation. Combine 2y and 43y to get 45y.

$$3x+45y=24$$

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.

$$93y+4x=23,45y+3x=24$$

To make 93y and 45y equal, multiply all terms on each side of the first equation by 45 and all terms on each side of the second by 93.

$$45\times 93y+45\times 4x=45\times 23,93\times 45y+93\times 3x=93\times 24$$

Simplify.

$$4185y+180x=1035,4185y+279x=2232$$

Subtract 4185y+279x=2232 from 4185y+180x=1035 by subtracting like terms on each side of the equal sign.

$$4185y-4185y+180x-279x=1035-2232$$

Add 4185y to -4185y. Terms 4185y and -4185y cancel out, leaving an equation with only one variable that can be solved.

$$180x-279x=1035-2232$$

Add 180x to -279x.

$$-99x=1035-2232$$

Add 1035 to -2232.

$$-99x=-1197$$

Divide both sides by -99.

$$x=\frac{133}{11}$$

Substitute \frac{133}{11} for x in 45y+3x=24. Because the resulting equation contains only one variable, you can solve for y directly.

$$45y+3\times \frac{133}{11}=24$$

Multiply 3 times \frac{133}{11}.

$$45y+\frac{399}{11}=24$$

Subtract \frac{399}{11} from both sides of the equation.

$$45y=-\frac{135}{11}$$

Divide both sides by 45.

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

The system is now solved.

$$y=-\frac{3}{11},x=\frac{133}{11}$$