Question

2x+3y=11and 2x-4y=24

Answer 1

Answer:

2x + 3y = 11 ---------------- (1)

2x - 4y = 24 ----------------(2)

now ,

by substituting method

equation 2 became ,

x = 2y + 12 --------- (divide by 2 ) ----- (3)

put equation (3) in equation (1).

equation 1 is

2 [ 2y + 12 ] + 3y = 11

4y + 24 + 3y = 11

7y + 24 = 11

7y = 11 - 24

7 y = - 13

y = -13/7

now put the value of y in in equation 3

x= 2 [-13/7 ] + 12

x= - 26 /7 + 12

x = -26 + 84 / 7

x = 58 /7

Answer 2

☯GIVEN :

• $\bf{2x +3y = 11}$ and $\bf{2x-4y = 24}$

☯TO DO :

• Solve by Elimination Method.

➲SOLUTION :

▪Multiply the first equation by 4 :

$\mapsto{ \sf{4(2x + 3y) = 11 \times 4}}$

$\mapsto{ \bf{8x+12y = 44 -(i)}}$

▪Multiply the second equation by 3 :

$\mapsto{ \sf {3(2x -4y) = 24\times 3}}$

$\mapsto{ \bf{6x-12y = 72- (ii)}}$

★Adding both equations:

$:\implies \sf(8x + 12y) +(6x -12y) = 44+72$

$:\implies \sf 8x + 12y +6x -12y = 116$

$:\implies \sf8x + 6x + \cancel{12y } - \cancel{12y }= 116$

$:\implies \sf 8x+6x = 116$

$:\implies \sf 14x = 116$

$:\implies \sf x = \cancel{\dfrac{116}{14}}$

$:\implies \huge{ \underline{\boxed{ \pink{\bf {x = \dfrac{58}{7} }}}}}$

✞Substituting the obtained value of x in the first equation :

$:\implies{ \sf {2x +3y = 11}}$

$:\implies {\sf{ 2\times \dfrac{58}{7} + 3y = 11}}$

$:\implies {\sf {\dfrac{116}{7}+3y = 11}}$

$:\implies{ \sf{ 3y = 11 - \dfrac{116}{7} }}$

$:\implies{ \sf{ 3y = \dfrac{77 - 116}{7} }}$

$:\implies {\sf{ 3y = \dfrac{ - 39}{7} }}$

$:\implies {\sf {y = \dfrac{ - 39}{7} \div 3}}$

$:\implies {\sf {y = \dfrac{ \cancel{-39}}{7} \times \dfrac{1}{ \cancel{3}} }}$

$:\implies\huge{ \underline{\boxed{ \blue{ \bf {y = \dfrac{ - 13}{7} }}}}}$

$\huge{\green{\therefore}}$ The value of x and y is $\bf{\dfrac{58}{7}}$ and $\bf{\dfrac{-13}{7}}$ respectively.