how do i solve 5(a + 3) + 4 = a - 1?

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stepherson's profile pic

stepherson | High School Teacher | In Training Educator

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The order of operations, PEMDAS, says to start with simplifying the parenthesis. It can be remembered with the phrase, “Please excuse my dear Aunt Sally.”

Parenthesis     Please

Exponents       Excuse

Multiply           My

Divide             Dear

Add                Aunt

Subtract         Sally

There is nothing to simplify inside the parenthesis. There are no exponents to simplify. There is a multiplication. We can multiply “5” times everything inside the parenthesis, a process known as “distibuting.”

`5(a + 3) + 4 = a - 1`

`5a + 15 + 4 = a - 1`

There is nothing more to multiply and nothing to divide. We can add `15 + 4`.

`5a + 15 + 4 = a - 1`

`5a + 19 = a - 1`

We have simplified both sides. Now we want our variable, `a` alone on one side and a number on the other side. We will make fewer mistakes if we can keep the variable positive. Let’s subtract `a` from both sides.

`5a - a + 19 = a - a - 1`

`4a + 19 = -1`

We now have our variable, `a` on one side, but it is not alone. Let’s subtract “19” from both sides.

`4a + 19 -19 = -1 - 19`

`4a = -20`

We have `4a` on one side and a number, `-20`  on the other. Since we want to solve for `a`  not `4a` let’s divide both sides by “4.”

`4a/4 = -20/4`

`a = -5`

t-rashmi's profile pic

t-rashmi | College Teacher | eNotes Newbie

Posted on

The given equation is

`5(a+3)+4=a-1`

We have to follow the order of operations to solve such mathematical equations. The BODMAS(Acronym for Bracket Of Division Multiplication Addition Subtraction) Rule defines which operations are to be performed first. According to this rule brackets are solved first.

So we first solve the bracket.Opening the bracket we get

`5*a+5*3+4=a-1`  Now we perform multiplication before addition. 

`5a+15+4=a-1`

`5a+19=a-1`

Subtracting 'a' from both sides of the equation we get,

`5a+19-a=a-1-a`

Rearranging to get all 'a's together for ease of addition and subtraction, we get

`5a-a+19=a-a-1`

`4a+19=-1`

Subtracting 19 from both sides of the equation we get

`4a+19-19=-1-19`

`4a=-20`

Dividing both sides of the equation by 4 we get

`(4a)/4 = -20/4`

`a=-5`

We can also verify our answer by substituting the value obtained into both sides of the equation.

Consider Left Hand Side (L.H.S.)

We have `5(a+3)+4` on the L.H.S. Substituting `a=-5` ,

LHS`=` `5(-5+3)+4`

      `=` `5(-2)+4`

      `=` `-10+4`

      `=-6`

Now consider Right Hand Side (R.H.S.)

We have a-1 on the R.H.S.. Substituting  `a=-5` ,

R.H.S.`=-5-1`

         `=-6`

We see that when we substitute a=-5 in the given equation `5(a+3)=a-1` we get L.H.S.= R.H.S. = -6

Hence the value `a =-5` obtained is correct.

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kspcr111 | In Training Educator

Posted on

`5(a + 3) + 4 = a - 1`

=>`5*a+ 5*3 +4 = a-1`

=>`5a +15 +4 = a-1`

=> `5a +19=a-1`

Arranging the variable a 's terms to the L.H.S and constants to R.H.S

=> `5a- a = -1 -19`

=> `4a = -20`

=> `a= -5`

is the solution of variable "a"

tdviola's profile pic

tdviola | In Training Educator

Posted on

Remember the order of operations: PEMDAS, which stands for:

Parenthesis, Exponents, Multiplication, Division, Addition, and Subtraction. 

This acronym reminds you of the order in which you do each operation that is required to simplify the equation. 

Also remember the concept of like terms: we can add 3 and 5, for instance, but not 3a and 5 because the variable a is only present in one term. 

Our equation is: 5(a+3) + 4 = a-1

First, we'll look at the parenthesis: (a+3). Since a and 3 are not like terms, there's nothing to be done here, so we can move on. 

There are no exponents in this equation, so we can move on. 

Multiplication: there is one multiplication in this equation. Remember that 5(a+3) = 5*(a+3). 

Using the Distributive Property, this operation yields 5*a + 5*3. This simplifies to 5a+15. 

We now have: 

5a+15 + 4 = a-1

Next, we have addition. We have two addition operations, but only one is between like terms: the 15+4. 

This gives us: 5a+19 = a-1. 

Now comes the part that is confusing for a lot of students. How do we isolate a?

Whenever we are looking to isolate a variable, our first step is to get everything that has an a on one side of the equals sign and everything that does not have an a on the other side.

In each case, we do this by "undoing" what is being done to that term, and repeating that action on the other side. Subtraction undoes addition and vice versa; multiplication undoes division and vice versa.

You can manipulate equations in any way you like, as long as you always do the exact same thing to both sides. 

In this case, I'm going to start by taking a from the right to the left by subtracting an a from both sides: 

5a+19-a = a-1-a

This simplifies after combining like terms to:

4a+19=-1

I'll now move the 19 to the other side in the same way: 

4a+19-19 = -1-19

4a = -20

Now, we divide both sides by 4 to isolate a: 

4a/4 = -20/4

a = -5 

We divided by 4 to "undo" the fact that 4 and a are multiplied together. 

Common mistakes to look out for in a problem such as this:

1. BE SURE to use the Distributive Property when you multiply 5 and (a+3). 

2. Be sure to always replicate any operation you perform to isolate a in the exact same way on the other side of the equation. 

3. Always double check everything when working with negative signs! "Dropping negatives" is very common so I highly recommend double checking all your steps and all your work. 

becamymica's profile pic

becamymica | High School Teacher | (Level 1) eNoter

Posted on

5(a+3) + 4 = a-1

expand brackets

5a + 15 + 4 = a-1

5a + 19 = a - 1

subtract a from both sides

4a + 19 = -1

subtract 19 from both sides

4a = -20

divide both sides by 4

a =-5

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