Solve {l}{a=2b}{a+4=3(c+4)}{a+b+c=78}

$$\left\{ \begin{array} { l } { a = 2 b } \\ { a + 4 = 3 ( c + 4 ) } \\ { a + b + c = 78 } \end{array} \right.$$

Solve for a, b, c

$$a=44$$

$$b=22$$

$$c=12$$

Short Steps Using Substitution

Substitute 2b for a in the second and third equation.

$$2b+4=3\left(c+4\right) 2b+b+c=78$$

Solve these equations for b and c respectively.

$$b=4+\frac{3}{2}c c=78-3b$$

Substitute 4+\frac{3}{2}c for b in the equation c=78-3b.

$$c=78-3\left(4+\frac{3}{2}c\right)$$

Solve c=78-3\left(4+\frac{3}{2}c\right) for c.

$$c=12$$

Substitute 12 for c in the equation b=4+\frac{3}{2}c.

$$b=4+\frac{3}{2}\times 12$$

Calculate b from b=4+\frac{3}{2}\times 12.

$$b=22$$

Substitute 22 for b in the equation a=2b.

$$a=2\times 22$$

Calculate a from a=2\times 22.

$$a=44$$

The system is now solved.

$$a=44 b=22 c=12$$