Solve {l}{x_{1}-x_{2}=a}{(3/4x_{1}-1)=a}{(-x_{2}+b_{2}=a}

$$\left. \begin{array} { l } { x _ { 1 } - x _ { 2 } = a } \\ { ( \frac { 3 } { 4 } x _ { 1 } - 1 ) = a } \\ { ( - x _ { 2 } + b _ { 2 } = a } \end{array} \right.$$

Solve for x_1, x_2

$$x_{1}=\frac{4\left(a+1\right)}{3}$$

$$x_{2}=\frac{a+4}{3}$$

Steps Using Substitution

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

$$\frac{3}{4}x_{1}-1=a,x_{1}-x_{2}=a$$

Pick one of the two equations which is more simple to solve for x_{1} by isolating x_{1} on the left hand side of the equal sign.

$$\frac{3}{4}x_{1}-1=a$$

Add 1 to both sides of the equation.

$$\frac{3}{4}x_{1}=a+1$$

Divide both sides of the equation by \frac{3}{4}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$x_{1}=\frac{4a+4}{3}$$

Substitute \frac{4a+4}{3} for x_{1} in the other equation, x_{1}-x_{2}=a.

$$\frac{4a+4}{3}-x_{2}=a$$

Subtract \frac{4+4a}{3} from both sides of the equation.

$$-x_{2}=\frac{-a-4}{3}$$

Divide both sides by -1.

$$x_{2}=\frac{a+4}{3}$$

The system is now solved.

$$x_{1}=\frac{4a+4}{3},x_{2}=\frac{a+4}{3}$$