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Solve for x, y (complex solution)
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Solve for x, y
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\left(3k+20\right)x=4c+3,5x+3y=c
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.
\left(3k+20\right)x=4c+3
Pick one of the two equations which is more simple to solve for x by isolating x on the left hand side of the equal sign.
x=\frac{4c+3}{3k+20}
Divide both sides by 3k+20.
5\times \frac{4c+3}{3k+20}+3y=c
Substitute \frac{4c+3}{3k+20} for x in the other equation, 5x+3y=c.
\frac{5\left(4c+3\right)}{3k+20}+3y=c
Multiply 5 times \frac{4c+3}{3k+20}.
3y=\frac{3\left(ck-5\right)}{3k+20}
Subtract \frac{5\left(4c+3\right)}{3k+20} from both sides of the equation.
y=\frac{ck-5}{3k+20}
Divide both sides by 3.
x=\frac{4c+3}{3k+20},y=\frac{ck-5}{3k+20}
The system is now solved.
\left(3k+20\right)x=4c+3,5x+3y=c
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.
\left(3k+20\right)x=4c+3
Pick one of the two equations which is more simple to solve for x by isolating x on the left hand side of the equal sign.
x=\frac{4c+3}{3k+20}
Divide both sides by 3k+20.
5\times \frac{4c+3}{3k+20}+3y=c
Substitute \frac{4c+3}{3k+20} for x in the other equation, 5x+3y=c.
\frac{5\left(4c+3\right)}{3k+20}+3y=c
Multiply 5 times \frac{4c+3}{3k+20}.
3y=\frac{3\left(ck-5\right)}{3k+20}
Subtract \frac{5\left(4c+3\right)}{3k+20} from both sides of the equation.
y=\frac{ck-5}{3k+20}
Divide both sides by 3.
x=\frac{4c+3}{3k+20},y=\frac{ck-5}{3k+20}
The system is now solved.