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How do you solve 3 linear equations with substitution?

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How do you solve 3 linear equations with substitution?

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The three lines x – y + 1 = 0, 2x + y – 4 = 0, and x + y + 5 = 0 intersect to form a triangle. Find the coordinates of the triangle. If they intersect to form a triangle, then they do not all intersect at one point. Instead, one pair intersect at one point, another pair at another point, and finally the last pair at another point. These three points make up a triangle. What one must do, then, is take the three possible pairs of simultaneous equations and solve for x and y in each of the three pairs separately. The solutions from these three sets of pairs will then yield the coordinates of the triangle that are required by the question. So, the three possible pairs are: x – y + 1 = 0 2x + y – 4 = 0 (Let’s call this pair “A”) 2x + y – 4 = 0 x + y + 5 = 0 (Let’s call this pair “B”) x – y + 1 = 0 x + y + 5 = 0 (Let’s call this pair “C”) Solving for x and y in A, B, and C provides us with the x and y coordinates of the triangle, ABC, So, let’s do it: First A x – y + 1 = 0 2x + y – 4 = 0 Add

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