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the sum of 4 times of one number in 3 times a second number is 65.The sum of the two numbers is 18. What is the larger of the two numbers

Answer :

[tex] \left \{ {{4x+3y=65\:(I)} \atop {x+y=18\:(II)}} \right. [/tex]
simplify by (-3) second equation
[tex] \left \{ {{4x+3y=65\:\:\:\:} \atop {x+y=18\:(-3)}} \right. [/tex]
[tex] \left \{ {{4x+\diagup\!\!\!\!\!3y=65} \atop {-3x-\diagup\!\!\!\!\!3y=-54}} \right. [/tex]
[tex] \left \{ {{4x=65} \atop {-3x=-54}} \right. [/tex]
[tex]\boxed{\boxed{x = 11}} \end{array}}\qquad\quad\checkmark[/tex]

Replace the found value of "x" in the second equation.
[tex]x+y=18\:(II)[/tex]
[tex]11 + y = 18[/tex]
[tex]y = 18 - 11[/tex]
[tex]\boxed{\boxed{y = 7}} \end{array}}\qquad\quad\checkmark[/tex]

Answer:
[tex]\boxed{\boxed{x\ \textgreater \ y}or\boxed{11\ \textgreater \ 7}} \end{array}}\qquad\quad\checkmark[/tex]

The larger of the two numbers be 11 or 7.

  • The calculation is as follows:

4x + 3y = 65

X + y = 18

here we used elimination, so we multiply the second equation by -3

4x + 3y = 65

-3X -3y = -54

X = 11

Now

X+y = 18

11 + y = 18

Y = 7

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