Question

Solve using cross multiplication:
4x+3y=17
3x-4y+6=0
I will mark as brainliest ​

Answer 1

Given equation are 4x + 3y = 17 and 3x - 4y + 6 = 0

Given equation are 4x + 3y = 17 and 3x - 4y + 6 = 0Comparing with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, We have

Given equation are 4x + 3y = 17 and 3x - 4y + 6 = 0Comparing with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, We havea1 = 4, b1 = 3, c1 = -17 and a2 = 3, b2 = - 4, c2 = 6

Given equation are 4x + 3y = 17 and 3x - 4y + 6 = 0Comparing with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, We havea1 = 4, b1 = 3, c1 = -17 and a2 = 3, b2 = - 4, c2 = 6Now, x =

Given equation are 4x + 3y = 17 and 3x - 4y + 6 = 0Comparing with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, We havea1 = 4, b1 = 3, c1 = -17 and a2 = 3, b2 = - 4, c2 = 6Now, x =

Given equation are 4x + 3y = 17 and 3x - 4y + 6 = 0Comparing with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, We havea1 = 4, b1 = 3, c1 = -17 and a2 = 3, b2 = - 4, c2 = 6Now, x = and

Given equation are 4x + 3y = 17 and 3x - 4y + 6 = 0Comparing with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, We havea1 = 4, b1 = 3, c1 = -17 and a2 = 3, b2 = - 4, c2 = 6Now, x = and

Given equation are 4x + 3y = 17 and 3x - 4y + 6 = 0Comparing with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, We havea1 = 4, b1 = 3, c1 = -17 and a2 = 3, b2 = - 4, c2 = 6Now, x = and b1c2-b2c1a1b2-a2b1 and y=c1a2-c2a1a1b2-a2b1

Given equation are 4x + 3y = 17 and 3x - 4y + 6 = 0Comparing with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, We havea1 = 4, b1 = 3, c1 = -17 and a2 = 3, b2 = - 4, c2 = 6Now, x = and b1c2-b2c1a1b2-a2b1 and y=c1a2-c2a1a1b2-a2b1 ⇒ x =

Given equation are 4x + 3y = 17 and 3x - 4y + 6 = 0Comparing with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, We havea1 = 4, b1 = 3, c1 = -17 and a2 = 3, b2 = - 4, c2 = 6Now, x = and b1c2-b2c1a1b2-a2b1 and y=c1a2-c2a1a1b2-a2b1 ⇒ x =

Given equation are 4x + 3y = 17 and 3x - 4y + 6 = 0Comparing with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, We havea1 = 4, b1 = 3, c1 = -17 and a2 = 3, b2 = - 4, c2 = 6Now, x = and b1c2-b2c1a1b2-a2b1 and y=c1a2-c2a1a1b2-a2b1 ⇒ x = and

Given equation are 4x + 3y = 17 and 3x - 4y + 6 = 0Comparing with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, We havea1 = 4, b1 = 3, c1 = -17 and a2 = 3, b2 = - 4, c2 = 6Now, x = and b1c2-b2c1a1b2-a2b1 and y=c1a2-c2a1a1b2-a2b1 ⇒ x = and

Given equation are 4x + 3y = 17 and 3x - 4y + 6 = 0Comparing with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, We havea1 = 4, b1 = 3, c1 = -17 and a2 = 3, b2 = - 4, c2 = 6Now, x = and b1c2-b2c1a1b2-a2b1 and y=c1a2-c2a1a1b2-a2b1 ⇒ x = and 3×6-(-4)×(-17)4×(-4)-3×3 and y=-17×3-6×44×(-4)-3×3

Given equation are 4x + 3y = 17 and 3x - 4y + 6 = 0Comparing with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, We havea1 = 4, b1 = 3, c1 = -17 and a2 = 3, b2 = - 4, c2 = 6Now, x = and b1c2-b2c1a1b2-a2b1 and y=c1a2-c2a1a1b2-a2b1 ⇒ x = and 3×6-(-4)×(-17)4×(-4)-3×3 and y=-17×3-6×44×(-4)-3×3⇒ x 25

25⇒ x = 2 and y = 3.

Answer 2

Answer:

We will apply the method of substitution

4x+3y=17

x=17-3y/4 mark it as equation 1

Put x value in 2nd equation

3(17-3y/4)-4y+6=0

51-9y/4-4y+6=0

Taking LCM

51-9y-16y+24=0

25y=75

y=3 Put this value in equation 1

17-3(3)/4=x

x=2