# help me plz.

• Aug 6th 2005, 05:44 PM
Hayabusa
help me plz.
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.
• Aug 6th 2005, 10:35 PM
rgep
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, 09:04 AM
Hayabusa
How did you solve that? I am sorry for being a pain in the butt. I just want to learn this because my book doesn't show me how to solve this type of problem. It shows it as an example but just gives you the answer in a graph form.
• Aug 7th 2005, 11:27 AM
rgep
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).