Question 147013
{{{5y + 3x = 10 }}}  Start with the second equation.



{{{3x+5y=10}}} Rearrange the terms.




Start with the given system of equations:

{{{system(7x-2y=4,3x+5y=10)}}}



{{{5(7x-2y)=5(4)}}} Multiply the both sides of the first equation by 5.



{{{35x-10y=20}}} Distribute and multiply.



{{{2(3x+5y)=2(10)}}} Multiply the both sides of the second equation by 2.



{{{6x+10y=20}}} Distribute and multiply.



So we have the new system of equations:

{{{system(35x-10y=20,6x+10y=20)}}}



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



{{{(35x-10y)+(6x+10y)=(20)+(20)}}}



{{{(35x+6x)+(-10y+10y)=20+20}}} Group like terms.



{{{41x+0y=40}}} Combine like terms. Notice how the y terms cancel out.



{{{41x=40}}} Simplify.



{{{x=(40)/(41)}}} Divide both sides by {{{41}}} to isolate {{{x}}}.



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{{{35x-10y=20}}} Now go back to the first equation.



{{{35(40/41)-10y=20}}} Plug in {{{x=40/41}}}.



{{{1400/41-10y=20}}} Multiply.



{{{41(1400/cross(41)-10y)=41(20)}}} Multiply both sides by the LCD {{{41}}} to clear any fractions.



{{{1400-410y=820}}} Distribute and multiply.



{{{-410y=820-1400}}} Subtract {{{1400}}} from both sides.



{{{-410y=-580}}} Combine like terms on the right side.



{{{y=(-580)/(-410)}}} Divide both sides by {{{-410}}} to isolate {{{y}}}.



{{{y=58/41}}} Reduce.



So our answer is {{{x=40/41}}} and {{{y=58/41}}}.



Which form the ordered pair *[Tex \LARGE \left(\frac{40}{41},\frac{58}{41}\right)].



This means that the two equations are consistent and independent.