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If a, b and c are real numbers such that [tex] \frac{b}{a} [/tex] =6 and [tex] \frac{b}{c} [/tex] = 3 . What is the value of [tex] \frac{a+b}{b+c} } [/tex] ? palagay po yun solution

Sagot :

We change the two equations:

[tex]\frac{b}{a} =6\\b=6a\\[/tex]

[tex] \frac{b}{c}=3\\b=3c\\[/tex]

[tex] \frac{a+b}{b+c} = \frac{a+6a}{3c+c} = \frac{7a}{4c} [/tex]

[tex] \frac{b}{a}= \frac{2b}{c} \\ \frac{1}{a}= \frac{2}{c} \\c=2a[/tex]

[tex] \frac{7a}{4c} = \frac{7a}{4(2a)} = \frac{7a}{8a}= \frac{7}{8} [/tex]


Find a:
[tex] \frac{b}{a} =6[/tex]
a = b/6

Find b:
[tex] \frac{b}{a} =6[/tex]
b = 6a

Find c:
[tex] \frac{b}{c} =3[/tex]

Since b = 2a:
[tex] \frac{2a}{c} = 3[/tex]

3c = 6a
3c/3 = 6a/3
c = 2a

[tex] \frac{a+b}{b+c} = \frac{a + 6a}{6a+2a} = \frac{7a}{8a}= \frac{7}{8} [/tex]

The answers is 7/8.