LINR 3| Lesson 1| Try This! Solving a System Solution

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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