Question 548422
{{{-8x+4y=-7}}}..........(1)
{{{4x-2y=-1}}}...........(2)
Multiply by -2 both sides of equation (2)
{{{-8x+4y=2}}}...........(3)

LHS of equation (1) and (3) are same but RHS of both equation is different. It means that values of x and y for both equation are different. Therefore it is impossible to know the value of x and y by solving these two equations.


{{{3x+7y=10}}}.............(1)
{{{5x=7-2y}}}
{{{5x+2y=7}}}..............(2)
Multiply (1) by 5 and (2) by 3
{{{15x+35y=50}}}.............(3)
{{{15x+6y=21}}}..............(4)
Subtract (4) from (3)
{{{15x+35y=50}}}.............(3)
{{{-15x-6y=-21}}}..............(4)
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{{{29y=29}}}
{{{29y/29=29}}}
{{{cross(29)y/cross(29)=cross(29)/cross(29)}}}
{{{y=1}}}
Put the value of y in (1)
{{{3x+7y=10}}}.............(1)
{{{3x+7(1)=10}}}
{{{3x+7=10}}}
{{{3x=10-7}}}
{{{3x=3}}}
{{{3x/3=3/3}}}
{{{x=1}}}
(x,y)=(1, 1)


Check
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Put the values of x and y in (1)
{{{3x+7y=10}}}.............(1)
{{{3(1)+7(1)=10}}}
{{{3+7=10}}}
{{{10=10}}}

Put the values of x and y in (2)
{{{5x+2y=7}}}..............(2)
{{{5(1)+2(1)=7}}}
{{{5+2=7}}}
{{{7=7}}}