Solve {l}{2x+0.05y=6400*2}{(4-2)x+(0.6-0.05)x=2(3.5-2)*6}

$$\left\{ \begin{array} { l } { 2 x + 0.05 y = 6400 \times 2 } \\ { ( 4 - 2 ) x + ( 0.6 - 0.05 ) x = 2 ( 3.5 - 2 ) \times 6 } \end{array} \right.$$

Solve for x, y

$$x = \frac{120}{17} = 7\frac{1}{17} \approx 7.058823529$$

$$y = \frac{4347200}{17} = 255717\frac{11}{17} \approx 255717.647058824$$

Steps Using Substitution

Consider the second equation. Subtract 2 from 4 to get 2.

$$2x+\left(0.6-0.05\right)x=2\left(3.5-2\right)\times 6$$

Subtract 0.05 from 0.6 to get 0.55.

$$2x+0.55x=2\left(3.5-2\right)\times 6$$

Combine 2x and 0.55x to get 2.55x.

$$2.55x=2\left(3.5-2\right)\times 6$$

Subtract 2 from 3.5 to get 1.5.

$$2.55x=2\times 1.5\times 6$$

Multiply 2 and 1.5 to get 3.

$$2.55x=3\times 6$$

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.

$$2.55x=3\times 6,2x+0.05y=6400\times 2$$

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

$$2.55x=3\times 6$$

Divide both sides of the equation by 2.55, which is the same as multiplying both sides by the reciprocal of the fraction.

$$x=\frac{120}{17}$$

Substitute \frac{120}{17} for x in the other equation, 2x+0.05y=6400\times 2.

$$2\times \frac{120}{17}+0.05y=6400\times 2$$

Multiply 2 times \frac{120}{17}.

$$\frac{240}{17}+0.05y=6400\times 2$$

Subtract \frac{240}{17} from both sides of the equation.

$$0.05y=\frac{217360}{17}$$

Multiply both sides by 20.

$$y=\frac{4347200}{17}$$

The system is now solved.

$$x=\frac{120}{17},y=\frac{4347200}{17}$$