$\frac{1}{2}x+\frac{1}{4}y=6 \ \ \ |\times 2 \\
\frac{1}{5}x+\frac{1}{3}y=1 \ \ \ |\times (-5) \\ \\
x+\frac{1}{2}y=12 \\
\underline{-x-\frac{5}{3}y=-5} \\
\frac{1}{2}y-\frac{5}{3}y=12-5 \\
\frac{3}{6}y-\frac{10}{6}y=7 \\
-\frac{7}{6}y=7 \\
y=7 \times (-\frac{6}{7}) \\
y=-6 \\ \\
x+\frac{1}{2}y=12 \\
x+\frac{1}{2} \times (-6)=12 \\
x-3=12 \\
x=12+3 \\
x=15 \\ \\
\boxed{(x,y)=(15,-6)}$

3 0

3 years ago

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kondor19780726 [428]

Answer:

b. infinite solutions

Step-by-step explanation:

$-\frac{1}{2}x+\frac{1}{2}y=-1\\\\\textsf{subtract} \ -\frac{1}{2}y \ \textsf{from both sides:}\\\\\implies -\frac{1}{2}x=-1-\frac{1}{2}y\\\\\textsf{mulitply both sides by -2:}\\\\\implies x=2+y\\\\\textsf{subtract} \ y \ \textsf{from both sides:} \ x-y=2$

Therefore, the graphs are the same.

If the graphs of the equations are the same, then there are an infinite number of solutions that are true for both equations.

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2 years ago

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