how can i teach word problems related to the linear equation equatio in one variable to the 6th grade?

Answer 1

Some children will be able to do them without any help. However, some children that age won't understand how do to story problems even with one variable. I think that I would show them how to form an equation from the problem.

I'm not really sure. There comes a point that the student just need to be able to understand abstract ideas to do Math. If a student can't do any abstract thought, it's very difficult to help them understand.

Answer 2

A fellow Sixth Grade Math Teacher!!! Anyway...

We all know how one variable equations are so important. But I would suggest that you decide what you want your students to be able to do to a one-step equation. Is it enough that they only solve adding or subtracting or will multiplication and division also be involved? Is the problem relevant to your students?

Money is always a great way to start, since the kids can tell you how much they need to buy 3 pieces of candy when they know the price of one, etc. I tend to teach one-step equations as a means to an end in solving a proportion. Since so much of the sixth grade year is spent with fractions and ratios, I simply add proportions so they can see what happens when you have an unknown in the mix.

Please check the source below for some great lessons in this area (The Barbie Bungee Cord is alot of fun), and e-mail if I can provide any further input.

Answer 3

ok, the first element to do is rewrite the sentences as equations. A is two years older than B: A = B + 2. F is two times as previous as A: F = 2A. B is two times as previous as S: B = 2S. the daddy is 40 years older than the sister: F = S + 40. So our equations are: A = B +2 F = 2A B = 2S F = S + 40 Now, if we may be able to rewrite 2 equations to have an identical 2 variables, then lets sparkling up them. attempt plugging the 0.33 one into the first and the second one into the fourth: A = 2S + 2 2A = S + 40 Now, plug the first of those 2 equatoins into the second one: 2(2S + 2) = S + 40 4S + 4 = S + 40 Subtract (S + 4) from both aspect: 3S = 36 Divide with suggestions from 3: S = 12. So the sister is 12. We already recognize A = 2S + 2, so A = 26. A is 26 years previous. If it concerns, B is 24 and F is fifty 2.

Answer 4

In teaching young children, physical examples are necessary. For example, for linear equations in a single variable, problems related to speed/time/distance would make sense.

Answer 5

One idea to help them stay interested is to use word problems that involve their names or topics they are interested in. Sometimes they will be able to figure out simple ones they can relate to in their heads and then you can show them how this is solved algebraically.

Answer 6

use real world problems.