\left\{ \begin{array}{l}{ 5 x + 3 y = c }\\{ ( 3 k + 20 ) x = 4 c + 3 }\end{array} \right.
Solve for x, y (complex solution)
\left\{\begin{matrix}x=\frac{4c+3}{3k+20}\text{, }y=-\frac{5-ck}{3k+20}\text{, }&k\neq -\frac{20}{3}\\x=\frac{3\left(-4y-1\right)}{20}\text{, }y\in \mathrm{C}\text{, }&c=-\frac{3}{4}\text{ and }k=-\frac{20}{3}\end{matrix}\right.
Solve for x, y
\left\{\begin{matrix}x=\frac{4c+3}{3k+20}\text{, }y=-\frac{5-ck}{3k+20}\text{, }&k\neq -\frac{20}{3}\\x=\frac{3\left(-4y-1\right)}{20}\text{, }y\in \mathrm{R}\text{, }&c=-\frac{3}{4}\text{ and }k=-\frac{20}{3}\end{matrix}\right.
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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.
Examples
Quadratic equation
{ x } ^ { 2 } - 4 x - 5 = 0
Trigonometry
4 \sin \theta \cos \theta = 2 \sin \theta
Linear equation
y = 3x + 4
Arithmetic
699 * 533
Matrix
\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]
Simultaneous equation
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
Limits
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}