Question 146532
Start with the given system of equations:

{{{system((2/5)x + (3/5)y = 16/5,4x + 2y = 40)}}}



{{{5((2/cross(5))x + (3/cross(5))y )=5( 16/cross(5))}}} Multiply the both sides of the first equation by the LCD 5.



{{{2x+3y=16}}} Distribute and multiply.



{{{-2(2x+3y)=-2(16)}}} Now multiply the both sides of the first equation by -2.



{{{-4x-6y=-32}}} Distribute and multiply.



So we have the new system of equations:

{{{system(-4x-6y=-32,4x+2y=40)}}}





Now add the equations together. You can do this by simply adding the two left sides and the two right sides separately like this:



{{{(-4x-6y)+(4x+2y)=(-32)+(40)}}}



{{{(-4x+4x)+(-6y+2y)=-32+40}}} Group like terms.



{{{0x+-4y=8}}} Combine like terms. Notice how the x terms cancel out.



{{{-4y=8}}} Simplify.



{{{y=(8)/(-4)}}} Divide both sides by {{{-4}}} to isolate {{{y}}}.



{{{y=-2}}} Reduce.



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{{{-4x-6y=-32}}} Now go back to the first equation.



{{{-4x-6(-2)=-32}}} Plug in {{{y=-2}}}.



{{{-4x+12=-32}}} Multiply.



{{{-4x=-32-12}}} Subtract {{{12}}} from both sides.



{{{-4x=-44}}} Combine like terms on the right side.



{{{x=(-44)/(-4)}}} Divide both sides by {{{-4}}} to isolate {{{x}}}.



{{{x=11}}} Reduce.



So our answer is {{{x=11}}} and {{{y=-2}}}.



Which form the ordered pair *[Tex \LARGE \left(11,-2\right)].



This means that the two equations are consistent and independent.



Notice when we graph the equations, we see that they intersect at *[Tex \LARGE \left(11,-2\right)]. So this visually verifies our answer.



{{{drawing(500,500,-3,21,-12,8,
grid(1),
graph(500,500,-3,21,-12,8,(16-2x)/(3),(40-4x)/(2)),
circle(11,-2,0.05),
circle(11,-2,0.08),
circle(11,-2,0.10)
)}}} Graph of {{{(2/5)x + (3/5)y = 16/5}}} (red) and {{{4x + 2y = 40}}} (green)