Solve {l}{150=9/4a+3/2b+c}{45=25a+5b+c}{225/2=4a+2b+c}

$$\left\{ \begin{array} { l } { 150 = \frac { 9 } { 4 } a + \frac { 3 } { 2 } b + c } \\ { 45 = 25 a + 5 b + c } \\ { \frac { 225 } { 2 } = 4 a + 2 b + c } \end{array} \right.$$

Solve for a, b, c

$$a=15$$

$$b = -\frac{255}{2} = -127\frac{1}{2} = -127.5$$

$$c = \frac{615}{2} = 307\frac{1}{2} = 307.5$$

Short Steps Using Substitution

Solve 150=\frac{9}{4}a+\frac{3}{2}b+c for c.

$$c=150-\frac{9}{4}a-\frac{3}{2}b$$

Substitute 150-\frac{9}{4}a-\frac{3}{2}b for c in the second and third equation.

$$45=25a+5b+150-\frac{9}{4}a-\frac{3}{2}b \frac{225}{2}=4a+2b+150-\frac{9}{4}a-\frac{3}{2}b$$

Solve these equations for b and a respectively.

$$b=-30-\frac{13}{2}a a=-\frac{150}{7}-\frac{2}{7}b$$

Substitute -30-\frac{13}{2}a for b in the equation a=-\frac{150}{7}-\frac{2}{7}b.

$$a=-\frac{150}{7}-\frac{2}{7}\left(-30-\frac{13}{2}a\right)$$

Solve a=-\frac{150}{7}-\frac{2}{7}\left(-30-\frac{13}{2}a\right) for a.

$$a=15$$

Substitute 15 for a in the equation b=-30-\frac{13}{2}a.

$$b=-30-\frac{13}{2}\times 15$$

Calculate b from b=-30-\frac{13}{2}\times 15.

$$b=-\frac{255}{2}$$

Substitute -\frac{255}{2} for b and 15 for a in the equation c=150-\frac{9}{4}a-\frac{3}{2}b.

$$c=150-\frac{9}{4}\times 15-\frac{3}{2}\left(-\frac{255}{2}\right)$$

Calculate c from c=150-\frac{9}{4}\times 15-\frac{3}{2}\left(-\frac{255}{2}\right).

$$c=\frac{615}{2}$$

The system is now solved.

$$a=15 b=-\frac{255}{2} c=\frac{615}{2}$$