Solve {l}{4a+2b+c=3000}{a+b+c=1450}{2a+b-2c=0}

$$\left\{ \begin{array} { l } { 4 a + 2 b + c = 3000 } \\ { a + b + c = 1450 } \\ { 2 a + b - 2 c = 0 } \end{array} \right.$$

Solve for a, b, c

$$a=350$$

$$b=500$$

$$c=600$$

Short Steps Using Substitution

Solve 4a+2b+c=3000 for c.

$$c=3000-4a-2b$$

Substitute 3000-4a-2b for c in the second and third equation.

$$a+b+3000-4a-2b=1450 2a+b-2\left(3000-4a-2b\right)=0$$

Solve these equations for b and a respectively.

$$b=-3a+1550 a=600-\frac{1}{2}b$$

Substitute -3a+1550 for b in the equation a=600-\frac{1}{2}b.

$$a=600-\frac{1}{2}\left(-3a+1550\right)$$

Solve a=600-\frac{1}{2}\left(-3a+1550\right) for a.

$$a=350$$

Substitute 350 for a in the equation b=-3a+1550.

$$b=-3\times 350+1550$$

Calculate b from b=-3\times 350+1550.

$$b=500$$

Substitute 500 for b and 350 for a in the equation c=3000-4a-2b.

$$c=3000-4\times 350-2\times 500$$

Calculate c from c=3000-4\times 350-2\times 500.

$$c=600$$

The system is now solved.

$$a=350 b=500 c=600$$