Find the solution to the following system of equations using substitution: x-3y=2 2x+5y=15
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x-3y=2 , 2x+5y=15
x-3y=2
=> x=2+3y. eq1
2x+5y=15
=> 2(2+3y)+5y=15
=> 4+6y+5y=15
=> 11y=15-4
11y= 11
y= 1
putting the value of y in eq 1
x-3y=2
x-3×1=2
x=2+3
x=5.
so x=5 Ar y= 1
x-3y=2
=> x=2+3y. eq1
2x+5y=15
=> 2(2+3y)+5y=15
=> 4+6y+5y=15
=> 11y=15-4
11y= 11
y= 1
putting the value of y in eq 1
x-3y=2
x-3×1=2
x=2+3
x=5.
so x=5 Ar y= 1
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