Solve {l}{5x+(5i)y=100i}{(5i)x+(5-10i)y=60+80i}

$$\left\{ \begin{array} { l } { 5 x + ( 5 i ) y = 100 i } \\ { ( 5 i ) x + ( 5 - 10 i ) y = 60 + 80 i } \end{array} \right.$$

Solve for x, y

$$x=12+16i$$

$$y=4+12i$$

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.

$$5x+5iy=100i,5ix+\left(5-10i\right)y=60+80i$$

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

$$5x+5iy=100i$$

Subtract 5iy from both sides of the equation.

$$5x=-5iy+100i$$

Divide both sides by 5.

$$x=\frac{1}{5}\left(-5iy+100i\right)$$

Multiply \frac{1}{5} times -5iy+100i.

$$x=-iy+20i$$

Substitute -iy+20i for x in the other equation, 5ix+\left(5-10i\right)y=60+80i.

$$5i\left(-iy+20i\right)+\left(5-10i\right)y=60+80i$$

Multiply 5i times -iy+20i.

$$5y-100+\left(5-10i\right)y=60+80i$$

Add 5y to \left(5-10i\right)y.

$$\left(10-10i\right)y-100=60+80i$$

Add 100 to both sides of the equation.

$$\left(10-10i\right)y=160+80i$$

Divide both sides by 10-10i.

$$y=4+12i$$

Substitute 4+12i for y in x=-iy+20i. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-i\left(4+12i\right)+20i$$

Multiply -i times 4+12i.

$$x=12-4i+20i$$

Add 20i to 12-4i.

$$x=12+16i$$

The system is now solved.

$$x=12+16i,y=4+12i$$

Steps Using Matrices

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

$$5x+5iy=100i,5ix+\left(5-10i\right)y=60+80i$$

Write the equations in matrix form.

$$\left(\begin{matrix}5&5i\\5i&5-10i\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}100i\\60+80i\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}5&5i\\5i&5-10i\end{matrix}\right).

$$inverse(\left(\begin{matrix}5&5i\\5i&5-10i\end{matrix}\right))\left(\begin{matrix}5&5i\\5i&5-10i\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&5i\\5i&5-10i\end{matrix}\right))\left(\begin{matrix}100i\\60+80i\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&5i\\5i&5-10i\end{matrix}\right))\left(\begin{matrix}100i\\60+80i\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&5i\\5i&5-10i\end{matrix}\right))\left(\begin{matrix}100i\\60+80i\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5-10i}{5\left(5-10i\right)-5i\times \left(5i\right)}&-\frac{5i}{5\left(5-10i\right)-5i\times \left(5i\right)}\\-\frac{5i}{5\left(5-10i\right)-5i\times \left(5i\right)}&\frac{5}{5\left(5-10i\right)-5i\times \left(5i\right)}\end{matrix}\right)\left(\begin{matrix}100i\\60+80i\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{20}-\frac{1}{20}i&\frac{1}{20}-\frac{1}{20}i\\\frac{1}{20}-\frac{1}{20}i&\frac{1}{20}+\frac{1}{20}i\end{matrix}\right)\left(\begin{matrix}100i\\60+80i\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\left(\frac{3}{20}-\frac{1}{20}i\right)\times \left(100i\right)+\left(\frac{1}{20}-\frac{1}{20}i\right)\left(60+80i\right)\\\left(\frac{1}{20}-\frac{1}{20}i\right)\times \left(100i\right)+\left(\frac{1}{20}+\frac{1}{20}i\right)\left(60+80i\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}12+16i\\4+12i\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=12+16i,y=4+12i$$

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.

$$5x+5iy=100i,5ix+\left(5-10i\right)y=60+80i$$

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

$$5i\times 5x+5i\times \left(5i\right)y=5i\times \left(100i\right),5\times \left(5i\right)x+5\left(5-10i\right)y=5\left(60+80i\right)$$

Simplify.

$$25ix-25y=-500,25ix+\left(25-50i\right)y=300+400i$$

Subtract 25ix+\left(25-50i\right)y=300+400i from 25ix-25y=-500 by subtracting like terms on each side of the equal sign.

$$25ix-25ix-25y+\left(-25+50i\right)y=-500+\left(-300-400i\right)$$

Add 25ix to -25ix. Terms 25ix and -25ix cancel out, leaving an equation with only one variable that can be solved.

$$-25y+\left(-25+50i\right)y=-500+\left(-300-400i\right)$$

Add -25y to \left(-25+50i\right)y.

$$\left(-50+50i\right)y=-500+\left(-300-400i\right)$$

Add -500 to -300-400i.

$$\left(-50+50i\right)y=-800-400i$$

Divide both sides by -50+50i.

$$y=4+12i$$

Substitute 4+12i for y in 5ix+\left(5-10i\right)y=60+80i. Because the resulting equation contains only one variable, you can solve for x directly.

$$5ix+\left(5-10i\right)\left(4+12i\right)=60+80i$$

Multiply 5-10i times 4+12i.

$$5ix+\left(140+20i\right)=60+80i$$

Subtract 140+20i from both sides of the equation.

$$5ix=-80+60i$$

Divide both sides by 5i.

$$x=12+16i$$

The system is now solved.

$$x=12+16i,y=4+12i$$