\left\{ \begin{array} { l } { - 2 = k + b } \\ { - 4 = 0 + b } \end{array} \right.
Solve for k, b
k=2
b=-4
Share
Copied to clipboard
-4=b
Consider the second equation. Anything plus zero gives itself.
b=-4
Swap sides so that all variable terms are on the left hand side.
-2=k-4
Consider the first equation. Insert the known values of variables into the equation.
k-4=-2
Swap sides so that all variable terms are on the left hand side.
k=-2+4
Add 4 to both sides.
k=2
Add -2 and 4 to get 2.
k=2 b=-4
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}