Answered

Find a value for y for which the expression (1-y)(3-y)(6-y) has each given value

i. -70

ii. 120

Sagot :

[tex]\\ \tt\Rrightarrow (1-y)(3-y)(6-y)=-70[/tex]

[tex]\\ \tt\Rrightarrow -y(1+1)(3+1)(6+1)=-70[/tex]

[tex]\\ \tt\Rrightarrow y(2)(4)(7)=70[/tex]

[tex]\\ \tt\Rrightarrow y(56)=70[/tex]

[tex]\\ \tt\Rrightarrow y=\dfrac{70}{56}[/tex]

[tex]\\ \tt\Rrightarrow y=\dfrac{5}{4}[/tex]

#2

[tex]\\ \tt\Rrightarrow (1-y)(3-y)(6-y)=120[/tex]

[tex]\\ \tt\Rrightarrow -y(1+1)(3+1)(6+1)=120[/tex]

[tex]\\ \tt\Rrightarrow -y(2)(4)(7)=120[/tex]

[tex]\\ \tt\Rrightarrow 56y=-120[/tex]

[tex]\\ \tt\Rrightarrow y=\dfrac{-120}{56}[/tex]

[tex]\\ \tt\Rrightarrow y=\dfrac{-15}{7}[/tex]

Mhanifaid

Answer:

  • i. y = 8, ii. y = -2

Step-by-step explanation:

  • (1 - y)(3 - y)(6 - y) = -70

We can see the divisors are 2 and 3 apart.

i.

Factorize -70 so that divisors are 2 and 3 apart:

  • - 70 = (- 2)*(- 5)*(- 7)

The smallest one is 1 - y and -7:

  • 1 - y = -7
  • y = 1 + 7
  • y = 8

ii

Factorize 120 so that divisors are 2 and 3 apart:

  • 120 = 3*5*8

The smallest one is 1 - y and 3:

  • 1 - y = 3
  • y = 1 - 3
  • y = -2

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