# Find the Numbers

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#### mathdad

##### Full Member
There are two numbers whose sum is 53. Three times the smaller number is equal to 19 more than the larger number. What are the numbers?

Set up:

Let x = large number
Let y = small number

x + y = 53...Equation A
3y = x + 19....Equation B

x + y = 53

y = 53 - x...Plug into B.

3(53 - x) = x + 19

159 - 3x = x + 19

-3x - x = 19 - 159

-4x = -140

x = -140/-4

x = 35...Plug into A or B.

I will use A.

35 + y = 53

y = 53 - 35

y = 18.

The numbers are 18 and 35.

Yes?

#### JeffM

##### Elite Member
Do the numbers satisfy both equation?

$$\displaystyle 35 + 18 = 53.$$ Checks.

$$\displaystyle 3 * 18 = 54 = 35 + 19.$$ Checks.

In algebra, you can always check your own MECHANICAL work, and you should. It avoids mistakes, builds confidence, is a necessary skill for taking tests, and, most importantly, is what you will need in any job that expects you to be able to do math.

#### Subhotosh Khan

##### Super Moderator
Staff member
There are two numbers whose sum is 53. Three times the smaller number is equal to 19 more than the larger number. What are the numbers?

Set up:

Let x = large number
Let y = small number

x + y = 53...Equation A
3y = x + 19....Equation B

x + y = 53

y = 53 - x...Plug into B.

3(53 - x) = x + 19

159 - 3x = x + 19

-3x - x = 19 - 159

-4x = -140

x = -140/-4

x = 35...Plug into A or B.

I will use A.

35 + y = 53

y = 53 - 35

y = 18.

The numbers are 18 and 35.

Yes?
When possible check your work. Most of the time that is a part of the process of solution.

There is a shorter way to accomplish the algebra/arithmetic part.

You have two equations,

x + y = 53...Equation A
3y = x + 19....Equation B

rewrite B to collect all the unknowns to LHS

x + y = 53...Equation A
3y - x = 19....Equation B'

Add A & B' (to eliminate 'x' from the equations) and get equation C

3y + y = 72....Equation C

4y = 72

y = 18

Use this value in equation 'A'

x + 18 = 53...Equation A

x = 53- 18 = 35

Now check your solution......

• mathdad

#### mathdad

##### Full Member
When possible check your work. Most of the time that is a part of the process of solution.

There is a shorter way to accomplish the algebra/arithmetic part.

You have two equations,

x + y = 53...Equation A
3y = x + 19....Equation B

rewrite B to collect all the unknowns to LHS

x + y = 53...Equation A
3y - x = 19....Equation B'

Add A & B' (to eliminate 'x' from the equations) and get equation C

3y + y = 72....Equation C

4y = 72

y = 18

Use this value in equation 'A'

x + 18 = 53...Equation A

x = 53- 18 = 35

Now check your solution......
What is wrong with my method?

#### Dr.Peterson

##### Elite Member
Nothing is wrong with your method. You used substitution, and did it correctly; Khan used addition, which can take just a little less writing than what you did, but is certainly not the only correct way, or even necessarily "better".

• mathdad

#### mathdad

##### Full Member
Nothing is wrong with your method. You used substitution, and did it correctly; Khan used addition, which can take just a little less writing than what you did, but is certainly not the only correct way, or even necessarily "better".
There are several methods for solving two equations in two variables, right? Matrix algebra is another useful tool.

#### Dr.Peterson

##### Elite Member
Correct.

In fact, each method can be applied to a given system of equations in several ways (which makes it interesting to grade tests). You can solve either equation for either variable and substitute, or eliminate either variable from the equations by adding, then get the other variable in a couple ways. And you can solve the matrix form by several different techniques. When there are three or more variables, it gets even better!

But still, solving the equations is the "easy" (routine) part, compared to setting them up from a word problem.

• mathdad

#### mathdad

##### Full Member
Correct.

In fact, each method can be applied to a given system of equations in several ways (which makes it interesting to grade tests). You can solve either equation for either variable and substitute, or eliminate either variable from the equations by adding, then get the other variable in a couple ways. And you can solve the matrix form by several different techniques. When there are three or more variables, it gets even better!

But still, solving the equations is the "easy" (routine) part, compared to setting them up from a word problem.
We can also graph two equations to see where they cross each other. The crossing point is the solution in the form (x, y).

#### Jomo

##### Elite Member
Mathdad,
I know that you can check these problems. Just admit that you like posting here.

#### mathdad

##### Full Member
Mathdad,
I know that you can check these problems. Just admit that you like posting here.
I joined the site to review math learned long ago.

#### Jomo

##### Elite Member
I joined the site to review math learned long ago.
No, you study from Sullivan to review math learned long ago

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