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factor [tex]-\frac{1}{3}[/tex] out of [tex]4-\frac{2}{3}y[/tex]

[tex]4-\frac{2}{3}y=-\frac{1}{3}[/tex] ( BLANK + BLANK [tex]y[/tex])


Sagot :

Answer:

[tex]\bold{-\frac{1}{3}(-12 + 2y)}[/tex]

Step-by-step explanation:

The original expression is

[tex]4 - \frac{2}{3}y[/tex]

Let us write the factored result as
[tex]-\frac{1}{3} (a + by)[/tex]

where a and b are constants to be determined

[tex]-\frac{1}{3} (a + by) = -\frac{1}{3} (a) -\frac{1}{3} (by) = -\frac{a}{3} + -\frac{by} {3}[/tex]

Matching like terms to the original equation we get

[tex]-\frac{a}{3} = 4[/tex]  

Multiplying by -3 on both sides gives

a = -12


Matching the y terms we get
[tex]-\frac{by}{3} = -\frac{2}{3}y[/tex] ==> [tex]-\frac{b}{3} = - \frac{2}{3}[/tex]  ==> b = 2

So the factored expression is

[tex]\bold{-\frac{1}{3}(-12 + 2y)}[/tex]

We can verify this if we multiply the terms inside the parentheses by [tex]-\frac{1}{3}[/tex]

[tex]-\frac{1}{3}(-12) =4[/tex]

[tex]-\frac{1}{3} (2y) = -\frac{2}{3}y[/tex]