Week 12 – PreCalc 11 – Aliessah's Blog

This week in PreCalc 11 we learned how to solve systems of equations algebraically. A system of equations is a collection of two or more equations with a same set of unknowns. In solving a system of equations, we try to find values for each of the unknowns that will satisfy every equation in the system. The systems can either be both linear, be both quadratic, or be one linear and one quadratic. The reason for finding the unknowns (typically x and y) is because it’s the point that, when both the equations are graphed, they intersect. If the $x = 10$ and $y = 3$ then the point the equations intersect when graphed is $(10, 3)$ But sometimes we can’t graph both the equations to find the solution(s). So to figure it out, we have to solve in algebraically by using a method called substitution.

Example:

$x + y = 5$ $y = (x + 1)^2$

Step 1: Isolate One Variable

The first step to solving a system is to make one of the two equation equal either one of the two variables. In this example, one of the equations is equal to y. To make things easier you can make the simpler equation equal one of the variables, too.

Step 2: Substitute

Now that we have one equation equal y we can place that equation into the other equation. At any place of the equation that I see y I’m going to place the expression that y equals

$x + y = 5$ $y = (x + 1)^2$

$x + (x + 1)^2 = 5$

Step 3: Solve for X

$x + (x + 1)(x +1) = 5$

$x + x^2 + 2x + 1 = 5$

$x^2 + 3x + 1 = 5$

$x^2 + 3x - 4 = 0$

$(x + 4)(x - 1) = 0$

$x = -4$ or $x = 1$

Step 4: Substitute

Now we have two x- coordinates that could lead to the solution. There is possible to have two solutions, but even if we have 2 x-coordinates we could only end up with 1 solution. With the 2 x-coordinates that we have, we put them back into the other equation and solve for y

$x + y = 5$