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If x + y = 4a/5, y + z = 7a/5 and z + x = 9a/5, then x + y + z =

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If x + y = 4a/5, y + z = 7a/5 and z + x = 9a/5, then x + y + z =  [#permalink]

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Originally posted by Bunuel on 01 Jun 2017, 12:51.
Last edited by SajjadAhmad on 07 Jul 2019, 01:22, edited 1 time in total.
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Re: If x + y = 4a/5, y + z = 7a/5 and z + x = 9a/5, then x + y + z =  [#permalink]

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New post 01 Jun 2017, 13:33
1
Given that
\(x + y = \frac{4a}{5}\), \(y + z = \frac{7a}{5}\) and \(z + x = \frac{9a}{5}\)

Adding all 3, we get
\(2 * (x + y + z) = \frac{4a}{5} + \frac{7a}{5} + \frac{9a}{5}\)
\(2 * (x + y + z) = \frac{(4a+7a+9a)}{5}\)
\(x + y + z = \frac{20a}{5}*\frac{1}{2}\)
\(x + y + z = 2a\) (Option C)
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Re: If x + y = 4a/5, y + z = 7a/5 and z + x = 9a/5, then x + y + z =  [#permalink]

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New post 02 Jun 2017, 04:52
Bunuel wrote:
If \(x + y = \frac{4a}{5}\), \(y + z = \frac{7a}{5}\) and \(z + x = \frac{9a}{5}\), then \(x + y + z =\)

(A) 7a/15
(B) a
(C) 2a
(D) 3a
(E) 4a

pushpitkc 's method is quick; it works well in this three-equation/three variable problem where each variable has the same coefficient.

Another way to solve: use two different equations to isolate one variable. This method is handy when the variables' coefficients in the different equations are not "nice" numbers and typically need manipulating.

Given

\(x + y = \frac{4a}{5}\)

Rewrite: 5x + 5y = 4a --------------(P)

\(y + z = \frac{7a}{5}\)

Rewrite: 5y + 5z = 7a --------------(Q)

\(z + x = \frac{9a}{5}\)

Rewrite 5x + 5z = 9a ---------------(R)

So x + y + z = ?

1. To eliminate x, and to get another equation with y and z, subtract (P) from (R):

5x + 5z = 9a
-5x - 5y = -4a
5z - 5y = 5a ----------------------------(S)

2. Add new equation (S), and (Q) from above:

5z - 5y = 5a
5z + 5y = 7a
10z + 0y = 12a, hence

z= \(\frac{12}{10}\)a ------------------------------------> z = \(\frac{6}{5}\)a

3. Find y

From original (Q) y + z = \(\frac{7a}{5}\)

y = \(\frac{7a}{5}\) - \(\frac{6a}{5}\) =\(\frac{a}{5}\)-------------------------> y = \(\frac{a}{5}\)

4. Find x

From original (P) x + y =\(\frac{4a}{5}\)

x = \(\frac{4a}{5}\) - \(\frac{a}{5}\) = \(\frac{3a}{5}\) ------------------------> x = \(\frac{3a}{5}\)

5. x + y + z = \(\frac{3a}{5}\) + \(\frac{a}{5}\) + \(\frac{6a}{5}\)

= \(\frac{10a}{5}\), or 2a

Answer C

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Re: If x + y = 4a/5, y + z = 7a/5 and z + x = 9a/5, then x + y + z =  [#permalink]

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New post 02 Jun 2017, 08:52
Adding all 3 gives 2(x+y+z)=20a/5----> x+y+z=2a
Answer C
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Re: If x + y = 4a/5, y + z = 7a/5 and z + x = 9a/5, then x + y + z =  [#permalink]

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New post 02 Jun 2017, 09:00
Bunuel wrote:
If \(x + y = \frac{4a}{5}\), \(y + z = \frac{7a}{5}\) and \(z + x = \frac{9a}{5}\), then \(x + y + z =\)

(A) 7a/15
(B) a
(C) 2a
(D) 3a
(E) 4a


\(( x + y ) + ( y + z ) + ( z + x ) = \frac{4a}{5} + \frac{7a}{5} + \frac{9a}{5}\)

Or, \(2x + 2y + 2z = \frac{4a+ 7a + 9a}{5}\)

Or, \(2 ( x + y + z ) = \frac{20a}{5}\)

Or, \(2 ( x + y + z ) = 4a\)

Or, \(( x + y + z ) = 2a\)

Hence, answer must be (C) 2a
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Re: If x + y = 4a/5, y + z = 7a/5 and z + x = 9a/5, then x + y + z =  [#permalink]

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New post 05 Jun 2017, 15:47
Bunuel wrote:
If \(x + y = \frac{4a}{5}\), \(y + z = \frac{7a}{5}\) and \(z + x = \frac{9a}{5}\), then \(x + y + z =\)

(A) 7a/15
(B) a
(C) 2a
(D) 3a
(E) 4a


Adding our 3 equations together, we get:

2x + 2y + 2z = 4a/5 + 7a/5 + 9a/5

2x + 2y + 2z = (4a + 7a + 9a)/5

2x + 2y + 2z = (20a)/5 = 4a

Dividing the entire equation by 2, we have:

x + y + z = 2a

Answer: C
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Re: If x + y = 4a/5, y + z = 7a/5 and z + x = 9a/5, then x + y + z =  [#permalink]

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Re: If x + y = 4a/5, y + z = 7a/5 and z + x = 9a/5, then x + y + z =   [#permalink] 24 Feb 2019, 13:40
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