Journal #1 – Ximena Turmel-2441bg6
Month: November 2018
Week 12 – Precalc 11
This week in precalculus 11, we learned how to solve systems algebraically. Solving algebraically includes substitution and elimination. In this blog post, I will be going over how to solve systems algebraically by substitution.
To substitute, you have to isolate the variable to one side of the equation. However, this is only done to one of the equations. When the variable is isolated, substitute it for the same variable on the other equation.
Ex.
x – 2y = -10 –> x = 2y – 10
3x – y = 0
3(2y – 10) – y = 0
Once it’s substitued into the other expression, there should only be one variable left in the ecpression. You then solve to find what the variable left in the espression is.
3(2y – 10) -y = 0
6y – 30 – y = 0
5y – 30 = 0
5y = 30
y = 6
Now that you’ve found y, you have to substitute the answer for y back into one of the equations after the variable has been isolated.
y = 6
x = 2y – 10
x = 2(6) – 10
x = 2
The solution is (2, 6).
Week 11 – Precalc 11
This week in precalculus 11, we learned how to graph inequalities in 2 variables. When graphing linear inequalities in 2 variables, you need a boundary line and you need to shade one side of the line.
Make sure you know that greater than (>) and less than (<) uses a dotted/broken boundary line and greater than or equal to (≥) and less than or equal to (≤) uses a solid boundary line.
When trying to solve an inequality in 2 variables, you have to isolate y.
Ex.
4x – 2y ≤ -6
-2y ≤ -4x – 6
*When dividing by a negative, you have to flip the sign.
y ≥ 2x + 3
Once y is isolated and you have the slope and y-intercept, you can sketch the graph.
Week 10 – Precalc 11
This week in Precalculus 11, we learned about quadratic inequalities and how to solve them. A quadratic inequality is when the equation is greater than 0 (x > 0), greater than or equal to 0 (x ≥ 0), less than 0 (x < 0), less than or equal to 0 (x ≤ 0), or equal to 0 (x = 0).
To solve a quadratic inequality, you have to make sure that one side is zero. If one side is not zero, then you have to move all the numbers to one side to make it zero.
Ex.
4 > x2 – 3x
0 > x2 – 3x – 4
Once all the terms are on one side and the other side is zero, you have to solve for x.
Ex.
0 > x2 – 3x – 4
0 > (x – 4)(x + 1)
x = 4 x = -1
A fast way to find the solution is to imagine a graph and to imagine the x-axis as the 0. We know that the x-intercepts are 4 and -1 and that a is positive, so the parabola will open up. Since 0 is greater than the x’s, we know that we have to find the inequalities for x when it’s above the x-axis. We can now establish that x < -1 and x > 4.
Week 9 – Precalc 11
This week in Precalc 11, we learned how to solve my modeling problems. This is solving a word problem by determining the relationships between the variables presented in the problem.
To solve these types of problems, you have to remember that quadratics work for 2 missing variables. Then, determine the variables in the problem and then any other information that the question may be asking. The questions will usually be using terminology like Sum, Product, Greatest, Least, etc. Next, identify the relationship between these variables. This will allow you to solve the rest of the problem. After the relationships are established, you can then solve to produce the answer you’re looking for.
