4.0 Student Essay Example
Simplifying Expressions
While working on this problem I learned a lot about basic properties of addition and multiplication, such as distributive property, inverse property, commutative properties, etc. What’s common about these problems is the order in which we are going to do the steps. The first step will always be to remove the parenthesis, which uses the distributive property. Second will always be combining like terms and adding related coefficients what we have been working on this week which is dealing with real numbers.
Here is the solution to the first problem:
2a(a-5)+ 4(a-5)
=2aa+2a(-5)+4a-4⋅5
=2a^2-10a+4a-20
=2a^2-6a-20
The steps used to simplify this equation are as follows:
First I remove all of the parentheses. This is called the distributive property. 2a(a-5)+ 4(a-5)=2aa+2a(-5)+4a-4⋅5
The second step is to figure out the correct coefficients:
2aa+2a(-5)+4a-4⋅5=2a^2-10a+4a-20
The final step is to find the like terms and then combine them to obtain the simplified form:
2a^2-10a+4a-20=2a^2-6a-20
Here is the solution to the second problem:
2w-3+3(w-4)-5(w-6)
=2w-3+3w-3⋅4-5w+5⋅6
=2w-3+3w-12-5w+30
=15
The steps used to simplify this particular equation are:
First of all the distributive property removes the parentheses.
2w-3+3(w-4)-5(w-6) =2w-3+3w-3⋅4-5w+5⋅6
The second step is finding the numbers in front of the letters. Those coefficients are -3*4 and 5*6. This looks like:
2w-3+3w-3⋅4-5w+5⋅6=2w-3+3w-12-5w+30
The next step is finding the like terms, e.g. 2w, 3w and 5w. Then I combined them into a simplified form:
2w-3+3w-12-5w+30=15
I need to point out that in order to reach the 15, I had to use the inverse property of addition which is 5w + (-5w) = 0
This brings me to the fully simplified answer of 15!
The final problem is:
0.05(0.3m+35n)-0.8(-0.09n-22m)
=0.05⋅0.3m+0.05⋅35n-0.8⋅(-0.09n)-0.8⋅(-22m)
=0.015m+1.75n+0.072n+17.6m
While working on this problem I learned a lot about basic properties of addition and multiplication, such as distributive property, inverse property, commutative properties, etc. What’s common about these problems is the order in which we are going to do the steps. The first step will always be to remove the parenthesis, which uses the distributive property. Second will always be combining like terms and adding related coefficients what we have been working on this week which is dealing with real numbers.
Here is the solution to the first problem:
2a(a-5)+ 4(a-5)
=2aa+2a(-5)+4a-4⋅5
=2a^2-10a+4a-20
=2a^2-6a-20
The steps used to simplify this equation are as follows:
First I remove all of the parentheses. This is called the distributive property. 2a(a-5)+ 4(a-5)=2aa+2a(-5)+4a-4⋅5
The second step is to figure out the correct coefficients:
2aa+2a(-5)+4a-4⋅5=2a^2-10a+4a-20
The final step is to find the like terms and then combine them to obtain the simplified form:
2a^2-10a+4a-20=2a^2-6a-20
Here is the solution to the second problem:
2w-3+3(w-4)-5(w-6)
=2w-3+3w-3⋅4-5w+5⋅6
=2w-3+3w-12-5w+30
=15
The steps used to simplify this particular equation are:
First of all the distributive property removes the parentheses.
2w-3+3(w-4)-5(w-6) =2w-3+3w-3⋅4-5w+5⋅6
The second step is finding the numbers in front of the letters. Those coefficients are -3*4 and 5*6. This looks like:
2w-3+3w-3⋅4-5w+5⋅6=2w-3+3w-12-5w+30
The next step is finding the like terms, e.g. 2w, 3w and 5w. Then I combined them into a simplified form:
2w-3+3w-12-5w+30=15
I need to point out that in order to reach the 15, I had to use the inverse property of addition which is 5w + (-5w) = 0
This brings me to the fully simplified answer of 15!
The final problem is:
0.05(0.3m+35n)-0.8(-0.09n-22m)
=0.05⋅0.3m+0.05⋅35n-0.8⋅(-0.09n)-0.8⋅(-22m)
=0.015m+1.75n+0.072n+17.6m