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Question
Solve, graphically, the following pairs of equations :
`(x + 1)/(4) = (2)/(3)(1 - 2y)`
`(2 + 5y)/(3) = x/(7) -2`
Solution
`(x + 1)/(4) = (2)/(3)(1 - 2y)`
⇒ `(x + 1)/(4)= (2)/(3) - (4y)/(3)`
⇒ 12 x `(x +1)/(4) =12 xx (2)/(3) - 12 xx (4y)/(3) `
⇒ 3(x + 1) = 8 - 16y
⇒ 3x + 3 = 8 - 16y
⇒ 3x + 3 - 8 = -16y
⇒ 3x - 5 = -16y
⇒ x = `(5 - 16y)/(3)`
The table for `(x+ 1)/(4) = (2)/(3)(1 - 2y)` is
X | 7 | - 9 | 23 |
Y | - 1 | 2 | - 4 |
Also, we have
`(2 + 5y)/(3) = x/(7) - 2`
⇒ 21 x `(2 + 5y)/(3) = 21 xx x/(7) - 21 xx 2`
⇒ 7(2 + 5y) = 3x - 42
⇒ 14 + 35y = 3x - 42
⇒ 3x = 14 + 35y + 42
⇒ 3x = 56 + 35y
⇒ x = `(56 + 35y)/(3)`
The table for `(2 + 5y)/(3) = x /(7) - 2` is
X | 7 | - 28 | 42 |
Y | - 1 | - 4 | 2 |
Plotting the points we get the following required graph:
From the above graph, it is dear that the two lines `(x + 1)/(4) = (2)/(3)(1 -2y)` and `(2 + 5y)/(3) = x /(7) - 2` intersect at the point (7, -1)
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