Solve {l}{46,2=c}{54,0=400a+20b+c}{57,9=1600a+40b+c}

$$\left\{ \begin{array} { l } { 46,2 = c } \\ { 54,0 = 400 a + 20 b + c } \\ { 57,9 = 1600 a + 40 b + c } \end{array} \right.$$

Solve for c, a, b

$$c=46,2$$

$$a=-0,004875$$

$$b=0,4875$$

Short Steps Using Substitution

Solve 46,2=c for c.

$$c=46,2$$

Substitute 46,2 for c in the second and third equation.

$$54=400a+20b+46,2 57,9=1600a+40b+46,2$$

Solve these equations for a and b respectively.

$$a=-\frac{1}{20}b+\frac{39}{2000} b=-40a+\frac{117}{400}$$

Substitute -\frac{1}{20}b+\frac{39}{2000} for a in the equation b=-40a+\frac{117}{400}.

$$b=-40\left(-\frac{1}{20}b+\frac{39}{2000}\right)+\frac{117}{400}$$

Solve b=-40\left(-\frac{1}{20}b+\frac{39}{2000}\right)+\frac{117}{400} for b.

$$b=\frac{39}{80}$$

Substitute \frac{39}{80} for b in the equation a=-\frac{1}{20}b+\frac{39}{2000}.

$$a=-\frac{1}{20}\times \frac{39}{80}+\frac{39}{2000}$$

Calculate a from a=-\frac{1}{20}\times \frac{39}{80}+\frac{39}{2000}.

$$a=-\frac{39}{8000}$$

The system is now solved.

$$c=46,2 a=-\frac{39}{8000} b=\frac{39}{80}$$