Today's soundtrack is Meat Loaf: Bat Out of Hell, an epic rock opera that sounds like it would make an incredible musical theatre production.
It's been a great day so far; my wife had some errands to run, and she took our youngest with her, so I took my kids out to get cheeseburgers, and then we went to the library and read books together. It was a really nice afternoon.
Now I'm going to be working on the assignment portion of the lesson I did yesterday on solving systems of equations algebraically.
As I mentioned yesterday, I'm going to go over some examples showing how the steps work.
I'm going to start with a linear-quadratic system.
My two equations are x - y = 2, and y = -x² + 10.
My first step while solving a linear-quadratic system of equations is to set up the equations: I need to put the quadratic equation into general form with y isolated, and I need the linear equation in the slope-intercept form, then I need to isolate x. My quadratic equation is already perfect, so I just needed to convert the linear equation to x = y + 2.
My two equations are now x = y + 2 and y = -x² + 10.
My second step is to put the linear equation into the quadratic equation, using substitution, which gives me y = -(y + 2)² + 10. I can break open the squared brackets: y = - (y+2) (y+2) +10, then I'll solve using distributive properties, giving me y = -y² - 4y -4 + 10, which simplifies to 0 = -y² - 5y + 6.
My two equations are now x = y + 2 and 0 = -y² - 5y + 6.
My third step is to solve the quadratic equation for y. I like using the quadratic formula to do so, but it can also be done using either decomposition of factors or finding the square. The two solutions that we get are y = -6 and y = 1. Now, we use substitution to put those values into our linear equation so that we can find our x-values. Since our linear equation is x = y + 2, and our y-values are -6 and 1, we first solve x = -6 + 2, which gives us -4. So the x-value of y -6 is -4; therefore, our first intersection between the two graphs is (-4, -6). Next, we do the same thing with the other y-value, which was 1: we solve x = 1 + 2, which comes to x = 3. Since we now know that the x-value of y = 1 is 3, our second intersection is (3, 1).
Our two solutions are (-4, -6) and (3, 1).
Finally, I used my graphing calculator to check my answer. I was able to successfully confirm that these are the two solutions.
Now, I'll try a quadratic-quadratic system.
My two equations are y = x² + 3x - 4 and y = 2x² - 5x + 2.
My first step while solving a quadratic-quadratic system of equations is to set up the equations. I need both equations in general form with y isolated, then I need to label the two equations: one will be called Equation 1, the other will be called Equation 2. Since my equations are both already in general form, all I need to do is label the two equations. I'll call the first Equation 1, and I'll call the second Equation 2.
My two equations are:
- Equation 1: y = x² + 3x - 4
- Equation 2: y = 2x² - 5x + 2
My second step is to put Equation 1 into Equation 2 by substituting Equation 1 for y, which gives me x² + 3x - 4 = 2x² - 5x + 2.
For the third step, I need to simplify the equation. By simplifying, I get 0 = x² - 8x + 6.
My two equations are:
- Equation 1: y = x² + 3x - 4
- Equation 2: 0 = x² - 8x + 6
The fourth step is to solve the second equation for x by factoring, completing the square, or using the quadratic formula. As I've said before, I'm partial to the quadratic formula. After solving the second equation, I get the two x-intercepts of the graph: 7.16 and 0.84.
The fifth step is to substitute both of the x-values into Equation 1 through substitution so that we can find their related y-values. Since my first x-intercept was 7.16, and my first equation is y = x² + 3x - 4, I'm going to use substitution to insert the number 7.16 anywhere that I see an x. This gives me y = 7.16² + 3(7.16) - 4. (7.16)(7.16) to the nearest hundredth is 51.27, and 7.16(3) is 21.48. 51.27 + 21.48 is 72.75; 72.75 minus 4 is 68.75. So the y-value of x 7.16 is 68.75, giving us our first solution: (7.16, 68.75). I now need to substitute our second x-value, 0.84, into Equation 1: y = 0.84² + 3(0.84) - 4. This gives me y = -0.77. So our second solution is (0.84, -0.77).
Our two solutions are (7.16, 68.75) and (0.84, -0.77)
Finally, using my graphing calculator, I confirmed that these are the correct solutions.
That's all for today!