Solve {l}{4a+2*b+2c=30}{a+b+c=10}{10a-7b+5c=27}

$$\left. \begin{array}{l}{ 4 a + 2 \cdot b + 2 c = 30 }\\{ a + b + c = 10 }\\{ 10 a - 7 b + 5 c = 27 }\end{array} \right.$$

Solve for a, b, c

$$a=5$$

$$b=4$$

$$c=1$$

Short Steps Using Substitution

Reorder the equations.

$$a+b+c=10 4a+2b+2c=30 10a-7b+5c=27$$

Solve a+b+c=10 for a.

$$a=-b-c+10$$

Substitute -b-c+10 for a in the second and third equation.

$$4\left(-b-c+10\right)+2b+2c=30 10\left(-b-c+10\right)-7b+5c=27$$

Solve these equations for b and c respectively.

$$b=5-c c=-\frac{17}{5}b+\frac{73}{5}$$

Substitute 5-c for b in the equation c=-\frac{17}{5}b+\frac{73}{5}.

$$c=-\frac{17}{5}\left(5-c\right)+\frac{73}{5}$$

Solve c=-\frac{17}{5}\left(5-c\right)+\frac{73}{5} for c.

$$c=1$$

Substitute 1 for c in the equation b=5-c.

$$b=5-1$$

Calculate b from b=5-1.

$$b=4$$

Substitute 4 for b and 1 for c in the equation a=-b-c+10.

$$a=-4-1+10$$

Calculate a from a=-4-1+10.

$$a=5$$

The system is now solved.

$$a=5 b=4 c=1$$