text: 125 views
  If $\textsf{x}$ and $\textsf{y}$ are non-negative integers such that $\textsf{x+9=z, y+1=z}$ and $\textsf{x+y<z+5},$ then the maximum possible value of $\textsf{2x+y}$ equals
  
  Given that, $x$ and $y$ are non-negative integers. That means  $x \geq 0, y \geq 0.$
  
  And,
  
  • $x + 9 = z \Rightarrow x = z – 9 \; \longrightarrow (1)$
  • $y + 1 = z \Rightarrow y = z – 1 \; \longrightarrow (2)$
  • $x + y < z + 5 \; \longrightarrow (3)$
  
  Put the value of $x,$ and $y$ in the equation $(3),$ we get.
  
  $x + y < z + 5$
  
  $\Rightarrow (z – 9) + (z – 1) < z + 5$
  
  $\Rightarrow z – 10 < 5$
  
  $\Rightarrow \boxed{z < 15}$
  
  Maximum value of $z$ can be $14.$
  
  So,
  
  • $x_{\textsf{max}} = 14 – 9 = 5$
  • $y_{\textsf{max}} = 14 – 1 = 13$
  
  Thus, the value of $2x+y = 2(5) + 13 = 23.$
  
  $\therefore$ The maximum possible value of $2x+y$ is $23.$
  
  Correct Answer$: 23$
  
  10.3k points
  
  1
  74 views
  1 vote
metadata: {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}