# Lesson 123 of Doodles Do Algebra

Today your child learns how to expand the solving of 2 unknowns using the various elimination techniques he learned last week. Today we do 3 unknowns and 3 equations.

It is essentially the same process as he learned last week. The only difference is that today he has to keep track of which equations he uses as he applies the elimination techniques (substitution, comparison, and addition and subtraction) and make sure to use all three.

If your child seems undaunted by 3 unknowns, let him go at it. There is no one right way of solving this, since you can get at the answer in any number of steps and by applying which ever of the elimination techniques. So if he works it all out then you are golden. If he gets stuck along the way, you need to tease out the point at which he made an error. If that is hard, just ask me in the comments below and I will help you out.

But if he is afraid to start the problem, just walk him through it yourself. Here is how you can do it:

First Step: use elimination by addition to add equation 2 to equation 3 (I chose this because there is a “-z” in equation 2 and a “z” in equation 3 so if you add them together, then the z is cancelled). Here is the result of the addition:

$x+y=15$

Second Step: solve that equation, we just got by elimination by addition in the first step, for x

$x=15-y$

Third Step: substitute the equation we got in the second step into the first original equation. That way we will have “touched” or used all three equations (very, very important) and we will be able to solve for y.

So substituting,

$2(15-y)+y=25$

$30-2y+y=25$

$y=5$

Fourth Step: now it really doesn’t matter which of the original equations you use next. Take the value you found for y (y=5) in the third step above, and substitute it into once of the other equations, we will choose equation 3.

$5+z=10$

$z=5$

Final Step: Now you know that y is 5 and z is 5. So we have a choice of which original equation to use to calculate x (either equation 1 or equation 2). I will choose equation 1 and substitute our calculated value for y and then solve for x.

$2x+5=25$

$2x=20$

$x=10$

Ta-Da!

# Lesson 122 – Doodles Do Algebra

Today your child practices solving another one of those “mental math” problems that he worked on at the beginning of the course only this time he can use the techniques for solving two independent equations that have 2 unknowns.

DoodlePig helps your child set up the problem, and it is kind of an easy one, so try not to let him just jump to the answer mentally, but if he does, rejoice that you have an intelligent child and move on.

1. A+100=B and B+100=2A

A=\$200 and B=\$300

# Lesson 121 – Doodles Do Algebra

Today your child learns the final method of solving for two variables when given two independent equations. This is called Elimination by Addition and Subtraction. The basic idea is to set up the equations so that you can either add or subtract them from (or to) each other and eliminate one of the variables. DoodlePoodle takes your child through an example and now that your child has is on her third method for solving this type of problem, she should have an easy time of it.

1. There is even a hint here to have your child try multiplying the first equation by a “2”. That gives a -2y term in the first equation and a 2y term in the second equation and when you add the two equations together, you get an equation that only contains x’s and then she can use the methods she learned for solving a simple equation with one variable to find the value of x…. and you get the idea by now. So x=4 and y=5.

# Lesson 120 – Doodles Do Algebra

Today your child learns the second way to solve for 2 unknowns in 2 independent equations. This is called Elimination by Comparison and basically what you do is manipulate each equation until you have one variable equaling a bunch of stuff that includes the second variable for both equations. Then you set the “bunch of stuff plus the second variable” side of both equations equal to each other. The basic idea is if a=n and a=p then n and p are equal.

DoodleCat walks your child through an example and then he gets two problems to try on his own.

1. x=2 and y=3

2. x=-36 and y=-40

# Lesson 119 – Doodles Do Algebra

Today your child starts the next step: solving simple equations for 2 unknowns.

The first point to notice is that in order to solve for 2 unknowns, you need to have 2 independent equations. You can remind your child that independent equations are ones that cannot be simply reduced to be equal. So x=4 and 2x=8 are not inependent since all you have to do is divide the second one by 2 on both sides and you have the same equation. If she is able and willing to engage in a broader discussion of independent equations, you can take the opportunity to show her that $(x-y)^2=3[math] is the same equation as [math]x^2-2xy+y^2=3$ because, as she saw in the course a month ago or so, you can factor the second equation into the first equation.

Ok, so now given that you have 2 independent equations, there are 3 ways to solve them. The first is called Elimination by Substitution and DoodleTwo explains the process very carefully on the worksheet. If this is complex for your child, just walk her through the first problem and then coax her into trying the second one solo. As always, don’t sorry if she won’t do the problem by herself at first. She will keep seeing these types of problems and eventually she will have the confidence to try one by herself. Afterall, you did not just toss her into the pool (I hope) to teach her to swim. You supported her and showed her and coaxed her until she learned the skills and felt confident enough to try them on her own. Why should math be any different?!

1. Step 1: solve the first equation for x. $x=-5y + 30$

Step 2: plug the value for x into the second equation $(y+10)/5-y/3=0$ and solve for y. y=15

Step 3: plug the value you just calculated for y back into one of the original equations and solve for x. And you get x=25

# Lesson 118 – Doodles Do Algebra

Today your child gets to do some word problems using what he learned about solving simple equations in the last several lessons. These are the same types of problems that he did when he began this Doodles Do Algebra series in the “mental math” section of the curriculum.

If you tell your child that he has already done problems like this – and done them in his head – he will have the confidence to overcome any initial uncertainty.

We walk your child through the solving process with boxes to fill in as a way to get him started setting up the problems.

1. x=first number and 2x (or twice the first) = second number. Then x+2x (or their sum) = 33. Now we just solve.

$x+2x=33$ first add the terms on the left since they both have the variable, x, in them

$3x=33$ now divide both sides by 3

$x=11$ and you have your answer: the first number is 11 and the second number is 22.

2. x=Daddy’s age and 2x (twice as old) = Grandpa’s age. Then 2x-x (or the difference between their ages) = 25. Now we solve for x.

$2x-x=25$ first combine the terms on the left since they both have the variable, x, in them

$x=25$ no dividing by the coefficient in this problem since the coefficient happens to be 1

Now you have your answer by looking at the original definitions: Daddy is 25 and Grandpa is 50.

# Lesson 117 – Doodles Do Algebra

Today is the last day of lessons on the topic of solving simple first degree equations. In the next lessons we will move onto the various methods of solving two unknowns, instead of just one.

But for today, your child needs positive feedback and encouragement. He needs to know that he has learned a tremendous amount of algebra so far and from now until the end of the course, it will only get more fun.

1. $x=abc/(c+b)$

2. $x=(b+c)/a$

3. $x=b^2/(b-a)$

# Lesson 116 – Doodles Do Algebra

Today’s lesson is yet more practice solving first degree equations and we add known quantities (like a, b, and c) into the mix along with the unknown and numbers. Every day we just add a bit more complexity to algebra and before you know it, your child will have learned the subject completely and without tears or frustration or those horrible feelings of inadequacy. Instead she will develop and add to her overall confidence and will have an open mind about learning seemingly difficult concepts in her future.

1. $x=(a-b)/2$

2. $x=(e-d)/(a+b-c)$

3. $x=ab/(-a^2+b^2)$

# Lesson 115 – Doodles Do Algebra

Today is more practice solving simple equations, just like yesterday’s lesson. Your child gets to work through problems that require all three of the steps in solving simple equations. We increase complexity of the problems as we go and look at what happens when one term in the equation is not a fraction, but the others are, and also how do you solve a simple first degree equation when there are fractional terms that include the unknown on both sides of the equation and they have different denominators. Just have your child follow the steps and he will see that they are really quite easy.

1. x= 30/11

2. x= 5

3. x= -35

# Lesson 114 – Doodles Do Algebra

In today’s lesson, DoodlePig (our fabulously intelligent guinea pig) teaches your child all three of the steps of solving a simple, first degree equation. She explains it really quite well (even though she is only a cartoon, and a cartoon guinea pig at that) so I don’t need to add anything to the teacher’s notes here.

DoodlePig works through two examples to get your child started.

Also, I wanted to note here that it really does not matter if an answer is written as x=a+b or x=b+a, or x=cb versus x=bc. The way your child chooses to write answers, if the content is correct, should not matter.