Question 133357
In order to solve this system

{{{16x^2 + 25y^2 = 400}}}

{{{5y + 4x = 20}}}


We need to isolate one variable and substitute that expression to eliminate one variable.



{{{5y + 4x = 20}}} Start with the second equation



{{{5y = 20-4x}}} Subtract 4x from both sides



{{{y = (20-4x)/5}}} Divide both sides by 5 to isolate y



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{{{16x^2 + 25y^2 = 400}}} Go back to the first equation


{{{16x^2 + 25((20-4x)/5)^2 = 400}}} Plug in {{{y = (20-4x)/5}}}. Notice how we eliminated the "y" terms and now have one equation with one unknown.



{{{16x^2 + 25((20-4x)^2/5^2) = 400}}} Distribute the outer exponent



{{{16x^2 + 25((20-4x)^2/25) = 400}}} Evaluate {{{5^2}}} to get 25



{{{16x^2 + cross(25)((20-4x)^2/cross(25)) = 400}}} Cancel like terms



{{{16x^2 + (20-4x)^2 = 400}}} Cancel like terms



{{{16x^2 + 400-160x+16x^2 = 400}}} Foil



{{{16x^2 + 400-160x+16x^2 - 400=0}}} Subtract 400 from both sides



{{{32x^2 -160x=0}}} Combine like terms



{{{32x(x -5)=0}}} Factor out {{{32x}}}




Now set each factor equal to zero:


{{{32x=0}}} or  {{{x -5=0}}} 


Now solve for x for each factor:


{{{x=0}}} or  {{{x=5}}} 




So our solutions are 


{{{x=0}}} or  {{{x=5}}} 


So let's find y when {{{x=0}}}


{{{5y + 4x = 20}}} Start with the second equation


{{{5y + 4(0) = 20}}} Plug in {{{x=0}}}


{{{5y = 20}}} Multiply and simplify



{{{y = 4}}} Divide both sides by 5 to isolate x



So when {{{x=0}}}, {{{y=4}}}



So one intersection point is (0,4)



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So let's find y when {{{x=5}}}


{{{5y + 4x = 20}}} Start with the second equation


{{{5y + 4(5) = 20}}} Plug in {{{x=0}}}


{{{5y +20= 20}}} Multiply and simplify



{{{5y =0}}} Subtract 20 from both sides



{{{y = 0}}} Divide both sides by 5 to isolate x



So when {{{x=5}}}, {{{y=0}}}



So one intersection point is (5,0)




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Answer:



So our solutions are  (0,4) and (5,0)





Notice if we graph, we can visually verify our answer



{{{ graph( 500, 500,  -10, 10, -10, 10, (4/5)*sqrt(-x^2+25),-(4/5)*sqrt(-x^2+25),-(4/5)*x+4) }}}