Question

If $x,y$ and $z$ are real numbers such that $4x+2y+z=31$ and $2x+4y-z=19,$ then the value of $9x+7y+z$

• A. equals $\frac{182}{3}$
• B. equals $\frac{218}{3}$
• C. equals $\frac{281}{3}$
• D. cannot be computed from the given information

Answer 1

The correct option is B equals $\frac{218}{3}$

Let $9x+7y+z=a(4x+2y+z)+b(2x+4y-z)$
Then $9x+7y+z=(4a+2b)x+(2a+4b)y+(a-b)z$
Comparing the coefficients, we get
$4a+2b=9\cdots \left(1\right)$
$2a+4b=7\cdots \left(2\right)$
$a-b=1\cdots \left(3\right)$
Solving $\left(1\right)$ and $\left(2\right),$
$a=\frac{11}{6},b=\frac{5}{6}$
$(\frac{11}{6},\frac{5}{6})$ also satisfies $\left(3\right).$

$\therefore 9x+7y+z=\frac{11}{6}(4x+2y+z)+\frac{5}{6}(2x+4y-z)$
$=\frac{11}{6}\times 31+\frac{5}{6}\times 19=\frac{218}{3}$