Determine which system goes with what

  a, c, b, none

Step-by-step explanation:

Since we're solving a number of simultaneous equations, and/or evaluating them for a variety of points, it makes a certain amount of sense to use a tool capable of simplifying the process.

The attached shows plots of the lines and their points of intersection. Each pair is plotted in the same color.

The solutions to the equations are ...

  (3, 7) -- a (red)

  (2, -1) -- c (green)

  (4, 14) -- b (blue)

  (4, 20) -- none (not on any line)

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You can also choose pairs by trying the points in the equations. Once we have settled on a solution, we don't need to try that set of equations anymore.

a with (3, 7): 7 = 4(3) -5 . . . . true

b with (2, -1): -1 = -7(2) +42 . . . . false

c with (2, -1): -1 +12(2) = 23 . . . . true

b with (4, 14): 14 = -7(4) +42 . . . . true

d with (4, 20): 2(20) +8(4) = 52 . . . . false

This tells us the likely answers are a, c, b, none.

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Additional comment

When the equation for a line is written in "general form", its expression is ...

  ax +by +c = 0

We know that a product will be zero if and only if one or both of the factors is zero. That lets us use the graphing calculator to plot both lines of a system of equations using a single equation.

Starting from (line 1 expression) = 0, and (line 2 expression) = 0, we form the product (line 1 expression)(line 2 expression) = 0. The graphing calculator does the rest, showing both lines, their x- and y-intercepts, and their point of intersection.

We used this method here simply for the purposes of (a) writing fewer equations, and (b) making both lines of a system the same color.

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Writing systems of equations in general form also has the benefit that it makes using the "cross multiplication method" of solution easy to use--in the event you want to solve the systems algebraically.

Here's more on the cross multiplication method of solving a pair of linear equations:

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step-by-step explanation:

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