☯GIVEN : [tex]\bf{3x +2y = 20}[/tex] and [tex]\bf{5x-3y = 8}[/tex] ☯TO DO : Solve by Elimination Method. ➲SOLUTION: ▪Multiply the first equation by 3 : [tex]\implies{ \sf3(3x + 2y) = 20 \times 3} \\ \\[/tex] [tex]\implies{ \bf9x +6y = 60 – (i)}[/tex] ▪Multiply the second equation by 2 : [tex]\implies{ \sf 2(5x – 3y) = 8 \times 2} \\ \\[/tex] [tex]\implies{ \bf 10x -6y = 16 – (ii)}[/tex] ▪Now, Adding both equations : [tex] \implies \sf(9x + 6y) + (10x – 6y) = 60 + 16 \\ \\[/tex] [tex]\implies \sf9x + 6y + 10x – 6y = 76 \\ \\ [/tex] [tex]\implies \sf9x + 10x + \cancel{6y }- \cancel{6y }= 76 \\ \\ [/tex] [tex]\implies \sf9x + 10x = 76\\ \\[/tex] [tex]\implies \sf19x = 76 \\ \\[/tex] [tex]\implies \sf x = \cancel{\dfrac{76}{19} } \\ \\ [/tex] [tex]\implies \huge{ \underline{\boxed{ \purple{\tt {x = 4 }}}}} [/tex] ✞Substituting the obtained value of x in the first equation : [tex]\implies \sf3x + 2y = 20 \\ \\ [/tex] [tex]\implies \sf3 \times 4 + 2y = 20 \\ \\[/tex] [tex]\implies \sf12 + 2y = 20 \\ \\ [/tex] [tex]\implies \sf 2y = 20 – 12 \\ \\ [/tex] [tex]\implies \sf 2y = 8 \\ \\ [/tex] [tex]\implies \sf y = \cancel{\dfrac{8}{2}} \\ \\[/tex] [tex]\implies\huge{ \underline{\boxed{ \purple{ \tt {y = 4 }}}}}[/tex] [tex]\huge{\green{\therefore}}[/tex] The value of x and y is 4 and 4 respectively. Reply
Given : 3x +2y = 20 5x-3y =8 To Find : Solve by Elimination Method Solution: 3x +2y = 20 Eq1 5x-3y =8 Eq2 3 * Eq1 + 2 * Eq2 will eliminate y => 9x + 6y + 10x – 6y = 60 + 16 => 19x = 76 => x = 4 5 * Eq1 – 3 * Eq2 will eliminate x =>15x + 10y – 15x +9y = 100 – 24 => 19y = 76 => y = 4 Hence x = y = 4 Learn More: solve X + Y = 83 and 4 x + Y = 182 by elimination method, Letv … brainly.in/question/17301814 solve the following by elimination method: x+y/3 = 0.9 ; 11 / ( x + y/2 … brainly.in/question/10926830 Reply
☯GIVEN :
☯TO DO :
➲SOLUTION:
▪Multiply the first equation by 3 :
[tex]\implies{ \sf3(3x + 2y) = 20 \times 3} \\ \\[/tex]
[tex]\implies{ \bf9x +6y = 60 – (i)}[/tex]
▪Multiply the second equation by 2 :
[tex]\implies{ \sf 2(5x – 3y) = 8 \times 2} \\ \\[/tex]
[tex]\implies{ \bf 10x -6y = 16 – (ii)}[/tex]
▪Now, Adding both equations :
[tex] \implies \sf(9x + 6y) + (10x – 6y) = 60 + 16 \\ \\[/tex]
[tex]\implies \sf9x + 6y + 10x – 6y = 76 \\ \\ [/tex]
[tex]\implies \sf9x + 10x + \cancel{6y }- \cancel{6y }= 76 \\ \\ [/tex]
[tex]\implies \sf9x + 10x = 76\\ \\[/tex]
[tex]\implies \sf19x = 76 \\ \\[/tex]
[tex]\implies \sf x = \cancel{\dfrac{76}{19} } \\ \\ [/tex]
[tex]\implies \huge{ \underline{\boxed{ \purple{\tt {x = 4 }}}}} [/tex]
✞Substituting the obtained value of x in the first equation :
[tex]\implies \sf3x + 2y = 20 \\ \\ [/tex]
[tex]\implies \sf3 \times 4 + 2y = 20 \\ \\[/tex]
[tex]\implies \sf12 + 2y = 20 \\ \\ [/tex]
[tex]\implies \sf 2y = 20 – 12 \\ \\ [/tex]
[tex]\implies \sf 2y = 8 \\ \\ [/tex]
[tex]\implies \sf y = \cancel{\dfrac{8}{2}} \\ \\[/tex]
[tex]\implies\huge{ \underline{\boxed{ \purple{ \tt {y = 4 }}}}}[/tex]
[tex]\huge{\green{\therefore}}[/tex] The value of x and y is 4 and 4 respectively.
Given : 3x +2y = 20
5x-3y =8
To Find : Solve by Elimination Method
Solution:
3x +2y = 20 Eq1
5x-3y =8 Eq2
3 * Eq1 + 2 * Eq2 will eliminate y
=> 9x + 6y + 10x – 6y = 60 + 16
=> 19x = 76
=> x = 4
5 * Eq1 – 3 * Eq2 will eliminate x
=>15x + 10y – 15x +9y = 100 – 24
=> 19y = 76
=> y = 4
Hence x = y = 4
Learn More:
solve X + Y = 83 and 4 x + Y = 182 by elimination method, Letv …
brainly.in/question/17301814
solve the following by elimination method: x+y/3 = 0.9 ; 11 / ( x + y/2 …
brainly.in/question/10926830