Question 1089420
{{{-x + y = 5}}}.....eq.1 
{{{4x + y = 10}}}.........eq2
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if you want to use substitution, start with eq.1 and solve it for {{{x}}} or {{{y}}}

{{{-x + y = 5}}}.....eq.1 I will solve it or {{{y}}}

{{{-x +x+ y = x+5}}}

{{{y = x+5}}}.....substitute it in eq2


{{{4x + (x+5) = 10}}}.........eq2...solve {{{x}}}

{{{4x + x+5 = 10}}}

{{{5x = 10-5}}}

{{{5x = 5}}}

{{{highlight(x = 1)}}}

now, go to {{{y = x+5}}}.....substitute {{{x = 1}}} and solve {{{y}}}

 {{{y = 1+5}}}
 {{{highlight(y = 6)}}}