Question

HELP ME PLEASE!!

Answer 1

$\bold{\huge{\underline{ Solution }}}$

Given :-

• Here, We have given two equations that is ,
• $\sf{ x + 2y = 13\:\: and\:\: x + 5y = 28}$

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,

$\sf{ x + 2y = 13 ...eq(1)}$

$\sf{ x + 5y = 28 ...eq(2)}$

By solving eq(1) :-

$\sf{ x + 2y = 13 }$

$\sf{ x = 13 - 2y ...eq(3)}$

Subsitute eq(3) in eq(2) :-

$\sf{ (13 - 2y) + 5y = 28 }$

$\sf{ 13 - 2y + 5y = 28 }$

$\sf{ 13 + 3y = 28 }$

$\sf{ 3y = 28 - 13 }$

$\sf{ 3y = 28 - 13 }$

$\sf{ 3y = 15 }$

$\sf{ y = }{\sf{\dfrac{ 15}{3}}}$

$\sf{ y = }{\sf{\cancel{\dfrac{ 15}{3}}}}$

$\bold{ y = 5 }$

Thus, The value of y is 5

Now,

Subsitute the value of y in eq(1) :-

$\sf{ x + 2(5) = 13 }$

$\sf{ x + 10 = 13 }$

$\sf{ x = 13 - 10 }$

$\bold{ x = 3 }$

Hence, The value of x and y are 5 and 3 .

$\bold{\huge{\underline{ Let's \: Verify }}}$

Equation 1 :-

$\sf{ x + 2y = 13 }$

$\sf{ 3 + 2(5) = 13 }$

$\sf{ 3 + 10 = 13 }$

$\sf{ 13 = 13 }$

$\bold{ LHS = RHS }$

Equation 2 :-

$\sf{ x + 5y = 28 }$

$\sf{ 3 + 5(5) = 28 }$

$\sf{ 3 + 25 = 28 }$

$\sf{ 28 = 28 }$

$\bold{ LHS = RHS }$