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Sagot :
[tex]\frac{1}{2}x+\frac{1}{4}y=6 \ \ \ |\times 2 \\
\frac{1}{5}x+\frac{1}{3}y=1 \ \ \ |\times (-5) \\ \\
x+\frac{1}{2}y=12 \\
\underline{-x-\frac{5}{3}y=-5} \\
\frac{1}{2}y-\frac{5}{3}y=12-5 \\
\frac{3}{6}y-\frac{10}{6}y=7 \\
-\frac{7}{6}y=7 \\
y=7 \times (-\frac{6}{7}) \\
y=-6 \\ \\
x+\frac{1}{2}y=12 \\
x+\frac{1}{2} \times (-6)=12 \\
x-3=12 \\
x=12+3 \\
x=15 \\ \\
\boxed{(x,y)=(15,-6)}[/tex]
1st Step:
[tex]\frac { 1 }{ 2 } x+\frac { 1 }{ 4 } y=6\\ \\ 4\left( \frac { 1 }{ 2 } x+\frac { 1 }{ 4 } y \right) =6\cdot 4\\ \\ 2x+y=24\\ \\ y=24-2x[/tex]
2nd Step:
[tex]\frac { 1 }{ 5 } x+\frac { 1 }{ 3 } y=1\\ \\ 15\left( \frac { 1 }{ 5 } x+\frac { 1 }{ 3 } y \right) =1\cdot 15[/tex]
[tex]\\ \\ 3x+5y=15\\ \\ 5y=15-3x\\ \\ 5y=3\left( 5-x \right) \\ \\ y=\frac { 3\left( 5-x \right) }{ 5 } [/tex]
3rd Step:
[tex]\frac { 3\left( 5-x \right) }{ 5 } =24-2x[/tex]
[tex]\\ \\ 5\cdot \frac { 3\left( 5-x \right) }{ 5 } =5\left( 24-2x \right) \\ \\ 3\left( 5-x \right) =120-10x[/tex]
[tex]\\ \\ 15-3x=120-10x\\ \\ -3x+10x=120-15\\ \\ 7x=105\\ \\ x=\frac { 105 }{ 7 }[/tex]
[tex]\\ \\ \therefore \quad x=15[/tex]
4th Step (When x=15):
[tex] y=24-2\left( 15 \right) \\ \\ y=24-30\\ \\ \therefore \quad y=-6[/tex]
[tex]\frac { 1 }{ 2 } x+\frac { 1 }{ 4 } y=6\\ \\ 4\left( \frac { 1 }{ 2 } x+\frac { 1 }{ 4 } y \right) =6\cdot 4\\ \\ 2x+y=24\\ \\ y=24-2x[/tex]
2nd Step:
[tex]\frac { 1 }{ 5 } x+\frac { 1 }{ 3 } y=1\\ \\ 15\left( \frac { 1 }{ 5 } x+\frac { 1 }{ 3 } y \right) =1\cdot 15[/tex]
[tex]\\ \\ 3x+5y=15\\ \\ 5y=15-3x\\ \\ 5y=3\left( 5-x \right) \\ \\ y=\frac { 3\left( 5-x \right) }{ 5 } [/tex]
3rd Step:
[tex]\frac { 3\left( 5-x \right) }{ 5 } =24-2x[/tex]
[tex]\\ \\ 5\cdot \frac { 3\left( 5-x \right) }{ 5 } =5\left( 24-2x \right) \\ \\ 3\left( 5-x \right) =120-10x[/tex]
[tex]\\ \\ 15-3x=120-10x\\ \\ -3x+10x=120-15\\ \\ 7x=105\\ \\ x=\frac { 105 }{ 7 }[/tex]
[tex]\\ \\ \therefore \quad x=15[/tex]
4th Step (When x=15):
[tex] y=24-2\left( 15 \right) \\ \\ y=24-30\\ \\ \therefore \quad y=-6[/tex]
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