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Complete the given ordered pairs so that each is a solution of the given equation.

D. x + y = 8
(?, 2) (?, -4) (-2, ?)

E. 5x = 30 - 3y
(9, ?) (?, 10) (6, ?)

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Sagot :

Zenzieeidn

LINEAR EQUATIONS

Answer:

D. Linear Pair (x + y = 8)

  1. (6,2)
  2. (12, -4)
  3. (-2, 10)

E. Linear Pair (5x = 30 - 3y)

  1. (9, -5)
  2. (0,10)
  3. (6,0)

Step-by-step explanation:

Substitute the given values in order to find the missing terms in the given linear equations.

D. Linear Pair (x + y = 8)

1. (_,2) y = 2

  • [tex]x + y = 8[/tex]

  • [tex]x + 2 = 8[/tex]

  • [tex]x = 8 - 2[/tex]

  • [tex]x = 6[/tex]

2. (_, -4) y = -4

  • [tex]x + y = 8[/tex]

  • [tex]x - 4 = 8[/tex]

  • [tex]x = 8 + 4[/tex]

  • [tex]x = 12[/tex]

3. (-2,_) x = -2

  • [tex]x + y = 8[/tex]

  • [tex] - 2 + y = 8[/tex]

  • [tex]y = 8 + 2[/tex]

  • [tex]y = 10[/tex]

E. Linear Pair (5x = 30 - 3y)

1. (9,_) x = 9

  • [tex]5x = 30 - 3y[/tex]

  • [tex]5(9) = 30 - 3y[/tex]

  • [tex]45 = 30 - 3y[/tex]

  • [tex]45 - 30 = - 3y[/tex]

  • [tex] \frac{15}{ - 3} = \frac{ \cancel{ - 3}y}{ \cancel{ - 3} }[/tex]

  • [tex] -5 = y[/tex]

2. (_,10) y = 10

  • [tex]5x = 30 - 3y[/tex]

  • [tex]5x = 30 - 3(10)[/tex]

  • [tex]5x = 30 - 30[/tex]

  • [tex] \frac{ \cancel5x}{ \cancel5} = \frac{0}{5} [/tex]

  • [tex]x = 0[/tex]

3. (6,_) x = 6

  • [tex]5x = 30 - 3y[/tex]

  • [tex]5(6) = 30 - 3y[/tex]

  • [tex]30 = 30 - 3y[/tex]

  • [tex]30 - 30 = - 3y[/tex]

  • [tex] \frac{0}{ - 3} = \frac{ \cancel{- 3y}}{ \cancel{ - 3} }[/tex]

  • [tex]0 = y[/tex]
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