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 = 7a10z + 0y = 12a, hence
z= \(\frac{12}{10}\)a ------------------------------------>
z = \(\frac{6}{5}\)a3. 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
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