Math, asked by kumbarmuttappa69, 15 days ago

solve the linear equation in two variables by substitution method root 3 x + root 2 y is equal to 6 and root 5 x + root 3 y is equal to 8 class 10th ​

Answers

Answered by hariharan81122
0

Answer:

Hi

Hi3x 2y = v3 -(1)

Hi3x 2y = v3 -(1)V5x+ 3y = v2--( 2)

Hi3x 2y = v3 -(1)V5x+ 3y = v2--( 2)multiply equation ( 1) with v3 and equation

Hi3x 2y = v3 -(1)V5x+ 3y = v2--( 2)multiply equation ( 1) with v3 and equation(2) with v2, we get,

Hi3x 2y = v3 -(1)V5x+ 3y = v2--( 2)multiply equation ( 1) with v3 and equation(2) with v2, we get,3x- V6y3-(3)

Hi3x 2y = v3 -(1)V5x+ 3y = v2--( 2)multiply equation ( 1) with v3 and equation(2) with v2, we get,3x- V6y3-(3)V10x + V6y = 2 -(4)

Hi3x 2y = v3 -(1)V5x+ 3y = v2--( 2)multiply equation ( 1) with v3 and equation(2) with v2, we get,3x- V6y3-(3)V10x + V6y = 2 -(4)add (3) and (4), we get

Hi3x 2y = v3 -(1)V5x+ 3y = v2--( 2)multiply equation ( 1) with v3 and equation(2) with v2, we get,3x- V6y3-(3)V10x + V6y = 2 -(4)add (3) and (4), we get3x + V10x = 5

Hi3x 2y = v3 -(1)V5x+ 3y = v2--( 2)multiply equation ( 1) with v3 and equation(2) with v2, we get,3x- V6y3-(3)V10x + V6y = 2 -(4)add (3) and (4), we get3x + V10x = 5x(3t 10) = 5

Hi3x 2y = v3 -(1)V5x+ 3y = v2--( 2)multiply equation ( 1) with v3 and equation(2) with v2, we get,3x- V6y3-(3)V10x + V6y = 2 -(4)add (3) and (4), we get3x + V10x = 5x(3t 10) = 5x = 5/(3+ V10)

Hi3x 2y = v3 -(1)V5x+ 3y = v2--( 2)multiply equation ( 1) with v3 and equation(2) with v2, we get,3x- V6y3-(3)V10x + V6y = 2 -(4)add (3) and (4), we get3x + V10x = 5x(3t 10) = 5x = 5/(3+ V10)x = [ 5( V10 3)( V10 +3 X V10 3)]

Hi3x 2y = v3 -(1)V5x+ 3y = v2--( 2)multiply equation ( 1) with v3 and equation(2) with v2, we get,3x- V6y3-(3)V10x + V6y = 2 -(4)add (3) and (4), we get3x + V10x = 5x(3t 10) = 5x = 5/(3+ V10)x = [ 5( V10 3)( V10 +3 X V10 3)]x= 5( 10-3y[(V10 - (3]

Hi3x 2y = v3 -(1)V5x+ 3y = v2--( 2)multiply equation ( 1) with v3 and equation(2) with v2, we get,3x- V6y3-(3)V10x + V6y = 2 -(4)add (3) and (4), we get3x + V10x = 5x(3t 10) = 5x = 5/(3+ V10)x = [ 5( V10 3)( V10 +3 X V10 3)]x= 5( 10-3y[(V10 - (3]x = 5( V10 -3 (10-9)

Hi3x 2y = v3 -(1)V5x+ 3y = v2--( 2)multiply equation ( 1) with v3 and equation(2) with v2, we get,3x- V6y3-(3)V10x + V6y = 2 -(4)add (3) and (4), we get3x + V10x = 5x(3t 10) = 5x = 5/(3+ V10)x = [ 5( V10 3)( V10 +3 X V10 3)]x= 5( 10-3y[(V10 - (3]x = 5( V10 -3 (10-9)x = 5V10 15

Hi3x 2y = v3 -(1)V5x+ 3y = v2--( 2)multiply equation ( 1) with v3 and equation(2) with v2, we get,3x- V6y3-(3)V10x + V6y = 2 -(4)add (3) and (4), we get3x + V10x = 5x(3t 10) = 5x = 5/(3+ V10)x = [ 5( V10 3)( V10 +3 X V10 3)]x= 5( 10-3y[(V10 - (3]x = 5( V10 -3 (10-9)x = 5V10 15putx value in equation (2), we get,

Hi3x 2y = v3 -(1)V5x+ 3y = v2--( 2)multiply equation ( 1) with v3 and equation(2) with v2, we get,3x- V6y3-(3)V10x + V6y = 2 -(4)add (3) and (4), we get3x + V10x = 5x(3t 10) = 5x = 5/(3+ V10)x = [ 5( V10 3)( V10 +3 X V10 3)]x= 5( 10-3y[(V10 - (3]x = 5( V10 -3 (10-9)x = 5V10 15putx value in equation (2), we get,V5 (5V10 - 15) + V3y= v2

Hi3x 2y = v3 -(1)V5x+ 3y = v2--( 2)multiply equation ( 1) with v3 and equation(2) with v2, we get,3x- V6y3-(3)V10x + V6y = 2 -(4)add (3) and (4), we get3x + V10x = 5x(3t 10) = 5x = 5/(3+ V10)x = [ 5( V10 3)( V10 +3 X V10 3)]x= 5( 10-3y[(V10 - (3]x = 5( V10 -3 (10-9)x = 5V10 15putx value in equation (2), we get,V5 (5V10 - 15) + V3y= v225v2 15V5+ v3y = v2

Hi3x 2y = v3 -(1)V5x+ 3y = v2--( 2)multiply equation ( 1) with v3 and equation(2) with v2, we get,3x- V6y3-(3)V10x + V6y = 2 -(4)add (3) and (4), we get3x + V10x = 5x(3t 10) = 5x = 5/(3+ V10)x = [ 5( V10 3)( V10 +3 X V10 3)]x= 5( 10-3y[(V10 - (3]x = 5( V10 -3 (10-9)x = 5V10 15putx value in equation (2), we get,V5 (5V10 - 15) + V3y= v225v2 15V5+ v3y = v2V3y = 2 - 25v2 +15V5

Hi3x 2y = v3 -(1)V5x+ 3y = v2--( 2)multiply equation ( 1) with v3 and equation(2) with v2, we get,3x- V6y3-(3)V10x + V6y = 2 -(4)add (3) and (4), we get3x + V10x = 5x(3t 10) = 5x = 5/(3+ V10)x = [ 5( V10 3)( V10 +3 X V10 3)]x= 5( 10-3y[(V10 - (3]x = 5( V10 -3 (10-9)x = 5V10 15putx value in equation (2), we get,V5 (5V10 - 15) + V3y= v225v2 15V5+ v3y = v2V3y = 2 - 25v2 +15V5y=(15V5 24v2 /v3

Hi3x 2y = v3 -(1)V5x+ 3y = v2--( 2)multiply equation ( 1) with v3 and equation(2) with v2, we get,3x- V6y3-(3)V10x + V6y = 2 -(4)add (3) and (4), we get3x + V10x = 5x(3t 10) = 5x = 5/(3+ V10)x = [ 5( V10 3)( V10 +3 X V10 3)]x= 5( 10-3y[(V10 - (3]x = 5( V10 -3 (10-9)x = 5V10 15putx value in equation (2), we get,V5 (5V10 - 15) + V3y= v225v2 15V5+ v3y = v2V3y = 2 - 25v2 +15V5y=(15V5 24v2 /v3y =[ 3(5v5-8 2) V( V3)

Hi3x 2y = v3 -(1)V5x+ 3y = v2--( 2)multiply equation ( 1) with v3 and equation(2) with v2, we get,3x- V6y3-(3)V10x + V6y = 2 -(4)add (3) and (4), we get3x + V10x = 5x(3t 10) = 5x = 5/(3+ V10)x = [ 5( V10 3)( V10 +3 X V10 3)]x= 5( 10-3y[(V10 - (3]x = 5( V10 -3 (10-9)x = 5V10 15putx value in equation (2), we get,V5 (5V10 - 15) + V3y= v225v2 15V5+ v3y = v2V3y = 2 - 25v2 +15V5y=(15V5 24v2 /v3y =[ 3(5v5-8 2) V( V3)y [ v3 x v3 (5V5-8v2 ( V3 )

x v3 (5V5-8v2 ( V3 )y [v3 (5/5 8v2) ]

x v3 (5V5-8v2 ( V3 )y [v3 (5/5 8v2) ]y 5V15- 8V6

I hope this helps you.

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