Solve the system of equations by substitution:

y = x^2 + 2x - 2 and y = 3x + 4

I tried working the problem out but I think I'm doing somthing wrong. I hope someone can help me.

Thank you in advance.

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- Aug 6th 2005, 06:44 PM #1

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- Aug 6th 2005, 11:35 PM #2
Since your equations are both of the form y = (something only involving x) then either can be used to substitute for y in the other and you'll get the same thing, namely an equation in x alone. In your example it would be x^2 + 2x - 2 = 3x + 4 which is a quadratic in x. You get two values of x, namely x=-2 and x=3 and the corresponding values of y are then -2 and 13. So the solutions are (x=-2,y=-2) and (x=3,y=13).

You might like to graph the two functions y=x^2+2x-2 and y=3x+4: one is a parabola and the other a straight line. They should intersect in two points, with coordinates giving you the two solutions.

- Aug 7th 2005, 10:04 AM #3

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- Aug 7th 2005, 12:27 PM #4
Solving by substitution applies to a system of equations where you can use one equation to express one of the variables in terms of the others.

Typically you'll encounter it it a pair of simultaneous equations such as 2x+3y = 8, 3x+5y = 13. Let's consider that example for a moment. Take one of the equations, say the first, and get y on its own on one side with the other side containing only x but not y. We have 3y = 8-2x, so y = (8-2x)/3. Now substitute this value for y into the second equation, so that 3x + 5(8-2x)/3 = 13. Multiplying up (by 3 in this case) we get 9x + 40 - 10x = 39, so that x=1. We said that y = (8-2x)/3 = (8-2)/3 = 2. Of course we now remember to check that x=1,y=2 satisfies both of the original equations.

Your example had two equations, each of the form y = (an expression involving only x). So it's easy to get y on its own, in fact it has already been done for you. Choose the first, say, y = x^2+2x-2 and substitute into y = 3x+4 to get x^2+2x-2 = 3x+4. Solve for x (two possible values as this is a quadratic equation) and each value of x gives you a value for y. Remember to check your answers.

The unusual thing in your example was that one of the equation involved an x^2 term. But fortunately each equation involved y only as a linear term (that is, y but not y^2 or any higher powers).