[tex]\bold{\huge{\underline{ Solution }}}[/tex]
Given :-
- Here, We have given two equations that is ,
- [tex]\sf{ x + 2y = 13\:\: and\:\: x + 5y = 28}[/tex]
To Find :-
- We have to find the value of x and y.
Concept used :-
- Linear equations are those equations which having highest power of degree as 1 .
- Linear equations can be solved by subsitute method, elimination method and cross multiplication method.
- Here, I have used substitution method to make calculation easier.
- In subsitute method, you have to subsitute the value of one variable of eq(1) in eq(2) .
Let's Begin :-
Here,
- We have two equations, let consider these two equations as eq(1) and eq(2) that is,
[tex]\sf{ x + 2y = 13 ...eq(1)}[/tex]
[tex]\sf{ x + 5y = 28 ...eq(2)}[/tex]
By solving eq(1) :-
[tex]\sf{ x + 2y = 13 }[/tex]
[tex]\sf{ x = 13 - 2y ...eq(3)}[/tex]
Subsitute eq(3) in eq(2) :-
[tex]\sf{ (13 - 2y) + 5y = 28 }[/tex]
[tex]\sf{ 13 - 2y + 5y = 28 }[/tex]
[tex]\sf{ 13 + 3y = 28 }[/tex]
[tex]\sf{ 3y = 28 - 13 }[/tex]
[tex]\sf{ 3y = 28 - 13 }[/tex]
[tex]\sf{ 3y = 15 }[/tex]
[tex]\sf{ y = }{\sf{\dfrac{ 15}{3}}}[/tex]
[tex]\sf{ y = }{\sf{\cancel{\dfrac{ 15}{3}}}}[/tex]
[tex]\bold{ y = 5 }[/tex]
Thus, The value of y is 5
Now,
Subsitute the value of y in eq(1) :-
[tex]\sf{ x + 2(5) = 13 }[/tex]
[tex]\sf{ x + 10 = 13 }[/tex]
[tex]\sf{ x = 13 - 10 }[/tex]
[tex]\bold{ x = 3 }[/tex]
Hence, The value of x and y are 5 and 3 .
[tex]\bold{\huge{\underline{ Let's \: Verify }}}[/tex]
Equation 1 :-
[tex]\sf{ x + 2y = 13 }[/tex]
[tex]\sf{ 3 + 2(5) = 13 }[/tex]
[tex]\sf{ 3 + 10 = 13 }[/tex]
[tex]\sf{ 13 = 13 }[/tex]
[tex]\bold{ LHS = RHS }[/tex]
Equation 2 :-
[tex]\sf{ x + 5y = 28 }[/tex]
[tex]\sf{ 3 + 5(5) = 28 }[/tex]
[tex]\sf{ 3 + 25 = 28 }[/tex]
[tex]\sf{ 28 = 28 }[/tex]
[tex]\bold{ LHS = RHS }[/tex]
