Question

by elimination method. 5x+3y=55 and 2x+4y=28​

Answer 1

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

➙5x+3y=55----×2

➙2x+4y= 28----×5

New Equation obtained

➙10x+6y= 110

➙10x+20y= 140

➙$(-)(-)\:(-)$

____________

-20y+6y= -140+110

-14y= -30

cancel (-) sign

14y= 30

y= 30/14=15/7

y= 15/7

Put y's value into this equation

➙5x+3y=55

➙5x+3×15/7=55

➙5x+45/7=55

➙5x= 55-45/7

➙5x= 385-45/7

➙5x= 340/7

➙x= 340/7×1/5

➙x=68/7

X= 68/7

Check:-

➙5x+3y= 55

Taking LHS

➙5(68/7)+3(15/7)

➙340/7+45/7

➙385/7

=55

Since,LHS = RHS(Verified)

Extra information=>

Elimination method is a way of eliminating one variable from equation of two variable so that we can find value of one variable.

Elimination means =>To Remove

Method step by step For Elimination Method :

step 1: Firstly Multiply with any non-zero value so that the coefficient of any one variable of both the terms be equal and gets eliminated.

Generally we multiply with coefficient of a variable of both equation to each term.

step2: After Put negative sign where value is positive and put positive sign where value is negative so that any one variable gets eliminated or subtracted and hence we find value of any one variable.

step3: Now,after finding value of one variable substitute it another equation.

Answer 2

The question is wrong as in eqn 1 it should be 35.