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Question
Use the substitution y = 3x + 1 to solve for x : 5(3x + 1 )2 + 6(3x + 1) – 8 = 0
Solution
y = 3x + 1
Now, 5(3x + 1)2 + 6(3x + 1) – 8 = 0
Substituting the value of 3x + 1, we get
5y2 + 6y - 8 = 0
⇒ 5y2 + 10y - 4y - 8 = 0 ...`{(∴5xx(-8) = -40),(∴ -40 = 10xx(-4)),(6 = 10 - 4):}}`
⇒ 5y(y + 2) -4(y + 2) = 0
⇒ (y + 2)(5y - 4) = 0
Either y + 2 = 0,
then y = -2
or
5y - 4 = 0,
then 5y = 4
⇒ y = `(4)/(5)`
(i) If y = -2, then
3x + 1 = -2
⇒ 3x = -2 - 1
⇒ 3x = -3
⇒ x = `(-3)/(3)`
= -1
(ii) If y = `(4)/(5)`, then
3x = 1 = `(4)/(5)`
⇒ 3x = `(4)/(5) - 1`
= `(-1)/(5)`
⇒ x = `(-1)/(5) xx (1)/(3)`
= `(-1)/(15)`
Hence x = -1, `(-1)/(15)`.
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