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Sagot :
Given the system of equations:
[tex]\begin{gathered} 16x+12y=336\text{ \lparen1\rparen} \\ 11x+15y=312\text{ \lparen2\rparen} \end{gathered}[/tex]The elimination method consists of multiplying either of the two equations in order to have the coefficients of x or y with the same value and be able to add or subtract.
In this case we are going to multiply the equation (2) by 16/11:
[tex]\begin{gathered} \frac{16}{11}(11x+15y)=\frac{16}{11}*(312) \\ 16x+\frac{16*15}{11}y=\frac{16*312}{11} \\ \\ 16x+\frac{240}{11}y=\frac{4992}{11}\text{ \lparen3\rparen} \end{gathered}[/tex]Now, subtracting (1) - (3)
[tex]\begin{gathered} 16x+12y=336 \\ - \\ 16x+\frac{240}{11}y=\frac{4992}{11} \end{gathered}[/tex]This is going to be eual to:
[tex](16x-16x)+(12y-\frac{240}{11}y)=(336-\frac{4992}{11})[/tex]Where: 16x-16x=0, therefore:
[tex]12y-\frac{240}{11}y=336-\frac{4992}{11}[/tex]Solving for y:
[tex]\begin{gathered} -\frac{108}{11}y=-\frac{1296}{11} \\ 108y=1296 \\ y=\frac{1296}{108}=12 \end{gathered}[/tex]Finally, replacing the value of y in (1) to find x:
[tex]\begin{gathered} 16x+12(12)=336 \\ 16x=336-144 \\ 16x=192 \\ x=\frac{192}{16}=12 \end{gathered}[/tex]Answer: the solution of the system is x=12, y=12.
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