Solve {l}{3a+4b=4lambda}{4lambda+3a=6}{3b+4lambda=3}

$$\left. \begin{array} { l } { 3 a + 4 b = 4 \lambda } \\ { 4 \lambda + 3 a = 6 } \\ { 3 b + 4 \lambda = 3 } \end{array} \right.$$

Solve for a, b, λ

$$a=1$$

$$b=0$$

$$\lambda =\frac{3}{4}=0.75$$

Short Steps Using Substitution

Solve 3a+4b=4\lambda for a.

$$a=-\frac{4}{3}b+\frac{4}{3}\lambda$$

Substitute -\frac{4}{3}b+\frac{4}{3}\lambda for a in the equation 4\lambda +3a=6.

$$4\lambda +3\left(-\frac{4}{3}b+\frac{4}{3}\lambda \right)=6$$

Solve the second equation for b and the third equation for \lambda .

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

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

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

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

$$\lambda =\frac{3}{4}$$

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

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

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

$$b=0$$

Substitute 0 for b and \frac{3}{4} for \lambda in the equation a=-\frac{4}{3}b+\frac{4}{3}\lambda .

$$a=-\frac{4}{3}\times 0+\frac{4}{3}\times \frac{3}{4}$$

Calculate a from a=-\frac{4}{3}\times 0+\frac{4}{3}\times \frac{3}{4}.

$$a=1$$

The system is now solved.

$$a=1 b=0 \lambda =\frac{3}{4}$$