There are two numbers whose sum is 50. Three times the first is 5 more than twice the second. What are the numbers?

Just carefully follow the words and turn them into math expressions and relationships. Since there are 2 numbers,


Let x = the first number
Let y = the second number

x + y = 50 since their sum is stated as 50. BTW, at this point we don't know which is the bigger number.
3x = 5 + 2y decode "is" as the = sign, and "5 more" as adding 5.

No solve these simply by addition by first rearranging the equations and multiplying the first by 2
2x + 2y = 100 doesn't change the equation since you multiplied both sides of the equal sign
3x - 2y = 5 to align the y terms so they drop out when both equations are added
5x = 105 and x = 105/5 = 21 then use the simple original equation (x + y = 50) to get y
3 + y = 50 so y = 50 - 21 = 29

Hence, x = 21 is the smaller number, and y = 29 are the 2 numbers that solve the problem

Well, these numbers sound like they're in a bit of a pickle! Let's see if we can help them out.

Let's call the first number "x" and the second number "y."

According to the problem, the sum of the two numbers is 50. So, we can write the equation:

x + y = 50

Now, it's time for some clown math!

The problem also states that three times the first number is 5 more than twice the second number. We can write this equation:

3x = 2y + 5

Now we have a system of equations! Time to put on our silly math caps and solve them:

Let's start by solving the first equation for x:
x = 50 - y

Now, plug this value of x into the second equation:
3(50 - y) = 2y + 5

A little more clown math magic:

150 - 3y = 2y + 5

Let's simplify:
150 - 5 = 2y + 3y
145 = 5y
y = 29

Now, let's find the value of x:
x = 50 - y
x = 50 - 29
x = 21

So, the two numbers are 21 and 29. Voila!

To find the two numbers, let's start by assigning variables to them.

Let's say the first number is x and the second number is y.

The problem tells us that their sum is 50, so we can write the equation: x + y = 50.

The problem also tells us that three times the first number (3x) is 5 more than twice the second number (2y). We can write this equation as: 3x = 2y + 5.

Now, we have a system of two equations:

Equation 1: x + y = 50
Equation 2: 3x = 2y + 5

To solve this system, we can use substitution or elimination method. Let's use the substitution method here.

From Equation 1, y = 50 - x.
We can substitute this expression for y in Equation 2, so we have: 3x = 2(50 - x) + 5.

Now, we can solve this equation for x:

3x = 100 - 2x + 5
3x + 2x = 100 + 5
5x = 105
x = 105/5
x = 21

Now that we have the value of x, we can substitute it back into Equation 1 to find y:

21 + y = 50
y = 50 - 21
y = 29

Therefore, the two numbers are 21 and 29.