Solve {l}{200=a+b+c}{150=a/4+1/2b+c}{150=9/4a+3/2b+c}

$$\left\{ \begin{array} { l } { 200 = a + b + c } \\ { 150 = \frac { a } { 4 } + \frac { 1 } { 2 } b + c } \\ { 150 = \frac { 9 } { 4 } a + \frac { 3 } { 2 } b + c } \end{array} \right.$$

Solve for a, b, c

$$a=-200$$

$$b=400$$

$$c=0$$

Short Steps Using Substitution

Multiply each equation by the least common multiple of denominators in it. Simplify.

$$200=a+b+c 600=a+2b+4c 600=9a+6b+4c$$

Solve 200=a+b+c for a.

$$a=-b+200-c$$

Substitute -b+200-c for a in the second and third equation.

$$600=-b+200-c+2b+4c 600=9\left(-b+200-c\right)+6b+4c$$

Solve these equations for b and c respectively.

$$b=-3c+400 c=240-\frac{3}{5}b$$

Substitute -3c+400 for b in the equation c=240-\frac{3}{5}b.

$$c=240-\frac{3}{5}\left(-3c+400\right)$$

Solve c=240-\frac{3}{5}\left(-3c+400\right) for c.

$$c=0$$

Substitute 0 for c in the equation b=-3c+400.

$$b=-3\times 0+400$$

Calculate b from b=-3\times 0+400.

$$b=400$$

Substitute 400 for b and 0 for c in the equation a=-b+200-c.

$$a=-400+200-0$$

Calculate a from a=-400+200-0.

$$a=-200$$

The system is now solved.

$$a=-200 b=400 c=0$$