Solve {l}{900+4000x=6*10^3y}{6*10^3y=3*10^3}

$$\left\{ \begin{array} { l } { 900 + 4000 x = 6 \times 10 ^ { 3 } y } \\ { 6 \times 10 ^ { 3 } y = 3 \times 10 ^ { 3 } } \end{array} \right.$$

Solve for x, y

$$x=\frac{21}{40}=0.525$$

$$y=\frac{1}{2}=0.5$$

Steps Using Substitution

Consider the second equation. Calculate 10 to the power of 3 and get 1000.

$$6\times 1000y=3\times 10^{3}$$

Calculate 10 to the power of 3 and get 1000.

$$6\times 1000y=3\times 1000$$

Consider the first equation. Calculate 10 to the power of 3 and get 1000.

$$900+4000x=6\times 1000y$$

Subtract 6\times 1000y from both sides.

$$900+4000x-6\times 1000y=0$$

Subtract 900 from both sides. Anything subtracted from zero gives its negation.

$$4000x-6\times 1000y=-900$$

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.

$$6\times 1000y=3\times 1000,\left(-6\times 1000\right)y+4000x=-900$$

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

$$6\times 1000y=3\times 1000$$

Divide both sides by 6000.

$$y=\frac{1}{2}$$

Substitute \frac{1}{2} for y in the other equation, \left(-6\times 1000\right)y+4000x=-900.

$$\left(-6\times 1000\right)\times \frac{1}{2}+4000x=-900$$

Multiply -6\times 1000 times \frac{1}{2}.

$$-3000+4000x=-900$$

Add 3000 to both sides of the equation.

$$4000x=2100$$

Divide both sides by 4000.

$$x=\frac{21}{40}$$

The system is now solved.

$$y=\frac{1}{2},x=\frac{21}{40}$$