1). a. Using elimination: \(2x + 3y = 11\) and \(x + 2y = 6\) , note that we could eliminate either \(x\) or \(y\). If we want to eliminate \(x\) we can multiply the second equation through by \(-2\) and adding them together.

\(2x + 3y = 11\)

\(-2x – 4y = -12\)

∴ \(-y = -1\) or \(y = 1\)

by substituting \(1\) for \(y\) in the top equation

\(2x + 3 = 11\) → \(x = 4\)

\((4, 1)\) is the point of intersection.

b. Using substitution: We can solve the second equation for \(x\); \(x = -2y + 6\)

\(2(-2y + 6) + 3y = 11\) →\(-4y +12 + 3y = 11\) ; \(y = 1\)

and \(2x + 3 = 11\); \(x = 4\). Intersection is at \((4, 1)\)

c. Did you get \((4,1)\) when graphing the equations?

2). \(x\) represents the cost of a ticket and \(y\) the cost of a container of popcorn.

\(4x +3y = 72\)

\(5x + 3y = 87\)

A ticket costs $\(15\) and a container of popcorn costs $\(4\).

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