Solve 8x+5y=9 and 3x+2y=4 by cross multiplication method

The given pair of linear equations is8x + 5y = 9    ...(i)3x + 2y = 4    ...(ii)

(I) By substitution method

From equation (ii), we have2y = 4 - 3x

So, the solution of the given pair of linear equations is x = -2,y = 5.

(II) By cross-multiplication methodLet us write the given pair of linear equations is8x + 5y - 9 = 0    ...(i)3x + 2y - 4 = 0    ... (ii)Solving the equations, we get

Given equation are 8x + 5y = 9 and 3x + 2y = 4
Comparing with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, We have

a1 = 8, b1 = 5, c1 = -9 and a2 = 3, b2 = 2, c2 = - 4

Now, x = `[ b_1c_2 - b_2c_1 ]/[ a_1b_2 - a_2b_1 ]  and  y = [ c_1a_2 - c_2a_1 ]/[ a_1b_2 - a_2b_1 ]` 

⇒ x = `[ 5 xx ( - 4 ) - 2 xx ( - 9 )]/[ 8 xx 2 - 3 xx 5 ]  and  y = [ - 9 xx 3 - ( - 4 ) xx 8 ]/[ 8 xx 2 - 3 xx 5 ]`

⇒ x = `[ - 20 + 18 ]/[ 16 - 15 ] and y = [ - 27 + 32 ]/[ 16 - 15 ]`

⇒ x = `-2/1 and y = 5/1`
⇒ x = - 2 and y = 5.

Page 2

Given equation are 4x - 3y - 11 = 0 and 6x + 7y - 5 = 0
Comparing with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, We have

a1 = 4, b1 = - 3, c1 = -11 and a2 = 6, b2 = 7, c2 = - 5

Now, x = `[ b_1c_2 - b_2c_1 ]/[ a_1b_2 - a_2b_1 ]  and  y = [ c_1a_2 - c_2a_1 ]/[ a_1b_2 - a_2b_1 ]` 

⇒ x = `[ - 3 xx ( - 5 ) - 7 xx ( - 11 )]/[ 4 xx 7 - 6 xx ( - 3 ) ]  and  y = [ - 11 xx 6 - ( - 5 ) xx 4 ]/[ 4 xx 7 - 6 xx ( - 3) ]`

⇒ x = `[ 15 + 77 ]/[ 28 + 18 ] and y = [ - 66 + 20 ]/[ 28 + 18 ]`

⇒ x = `92/46 and y = - 46/46`

⇒ x = 2 and y = - 1

Page 3

Given equation are 4x + 6y = 15 and 3x - 4y = 7
Comparing with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, We have

a1 = 4, b1 = 6, c1 = -15 and a2 = 3, b2 = - 4, c2 = -7

Now, x = `[ b_1c_2 - b_2c_1 ]/[ a_1b_2 - a_2b_1 ]  and  y = [ c_1a_2 - c_2a_1 ]/[ a_1b_2 - a_2b_1 ]` 

⇒ x = `[ 6 xx (-7) - ( - 4 ) xx ( - 15 )]/[ 4 xx (-4) - 3 xx 6 ]  and  y = [ - 15 xx 3 - ( - 7 ) xx 4 ]/[ 4 xx ( - 4 ) - 3 xx 6 ]`

⇒ x = `[ - 42 - 60 ]/[ -16 - 18]  and y = [ - 45 + 28 ]/[ - 16 - 18 ]`

⇒ x `(-102)/-34  and y = (-17)/(-34)`

⇒ x = 3 and y = `1/2`

Page 4

Given equation are 0.4x - 1.5y = 6.5 and 0.3x + 0.2y = 0.9
Comparing with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, We have

a1 = 0.4, b1 = -1.5, c1 = -6.5 and a2 = 0.3, b2 = 0.2, c2 = 0.9

Now, x = `[ b_1c_2 - b_2c_1 ]/[ a_1b_2 - a_2b_1 ]  and  y = [ c_1a_2 - c_2a_1 ]/[ a_1b_2 - a_2b_1 ]` 

⇒ x = `[( - 1.5) xx (-0.9) - ( 0.2 ) xx ( -6.5 )]/[ 0.4 xx (0.2) - (0.3) xx (-1.5) ]  and  y = [ (-6.5) xx (0.3) - (-0.9) xx (0.4) ]/[ 0.4 xx ( 0.2) - (0.3) xx (-1.5) ]`

⇒ x = `[ 1.35 + 1.3 ]/[ 0.08 + 0.45 ]  and y = [ -1.95 + 0.36 ]/[ 0.08 + 0.45 ]`

⇒ x = `2.65/0.53 and y = (-1.59)/(0.53)`

⇒ x = 5 and y = - 3

Page 5

Given equation are √2x - √3y = 0 and √5x + √2y = 0
Comparing with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, We have

a1 = √2, b1 = √3, c1 = 0 and a2 = √5, b2 = √2, c2 = 0

Now, x = `[ b_1c_2 - b_2c_1 ]/[ a_1b_2 - a_2b_1 ]  and  y = [ c_1a_2 - c_2a_1 ]/[ a_1b_2 - a_2b_1 ]` 

⇒ x = `[ (-sqrt3) xx 0 - sqrt2 xx 0 ]/[ sqrt2 xx sqrt2 - sqrt5 xx (-sqrt3) ]  and  y = [ 0 xx sqrt5 - 0 xx sqrt2 ]/[ sqrt2 xx sqrt2 - sqrt5 xx ( - sqrt3 ) ]`

⇒ x = `0/[ 2 + sqrt15 ] and y = 0/[ 2 + sqrt15 ]`

⇒ x = 0 and y = 0.