Chia683
5

# What are the x-values of the solutions of this system? $\left \{ {{y=x+3} \atop {y=2x^2-3}} \right.$

So, We notice that both equations say that "y" is equal to something.  Since y is constant, we can assume that both of the other sides are equal, since they are both equal to y. First, we can set the sides equal to each other. $x+3=2x^2-3$ Now we can solve for x. Subtract x from both sides. $3=2x^2-x-3$ Subtract 3 from both sides. $0=2x^2-x-6\ or\ 2x^2-x-6=0$ We can now factor. $2x^2-x-6=0--\ \textgreater \ (x-2)(2x+3)=0$ Set each factor equal to zero. x - 2 = 0 Add 2 to both sides. x = 2 2x + 3 = 0 Subtract 3 from both sides. 2x = -3 Divide both sides by 2. $x= \frac{-3}{2}$ We can now substitute 2 for x in the first equation and see what we get for y. y = 2 + 3 y = 5 So one pair of solutions is (2,5). Now, let's see what we get if we use the second value for x. $y= \frac{-3}{2} +3$ $y= \frac{-3}{2} + \frac{6}{2}$ $y= \frac{3}{2}$ So another ordered pair is $( \frac{-3}{2} , \frac{3}{2} )$ Let's check our solutions. (2,5) We already know this works in the first equation, so we need to test the second equation. $5=2(2)^2-3$ 5 = 2(4) - 3 5 = 8 - 3 5 = 5 This checks. Now, let's check the second solution. We already know this works in the first equation, as we used that equation to find the y value.  We just need to test the second equation. $\frac{3}{2} = 2(-\frac{3}{2}) ^2-3$ $\frac{3}{2} = 2( \frac{9}{4}) -3$ $\frac{3}{2} = \frac{9}{2} -3$ $\frac{3}{2} = \frac{9}{2} - \frac{6}{2}$ $\frac{3}{2} = \frac{3}{2}$ This also checks. Therefore, our solutions are: S = {$(2,5),( \frac{3}{2} ,- \frac{3}{2} )$}