Let t represent soft tacos, and B represent burritos. Make an equation for both orders, in any form.
7 answers
Let's define the variables:
Based on your and your friend's orders, we can create the following equations:
\[ 3t + 3b = 11.25 \]
\[ 4t + 2b = 10.00 \]
So, the equations for both orders are:
To solve for the cost of one soft taco (\( t \)) and one burrito (\( b \)), we can use either the substitution method or the elimination method. Here, I'll illustrate both methods, but the elimination method might be more straightforward in this case due to the coefficients.
We have the two equations:
To eliminate one of the variables, we can manipulate the equations. Let's first simplify Equation 1 by dividing everything by 3:
\[ t + b = \frac{11.25}{3} = 3.75 \] So, we rewrite Equation 1 as: \[ t + b = 3.75 \quad (Equation 3) \]
Next, we can manipulate Equation 2. To make the coefficients of \( b \) in both equations easy to eliminate, we can multiply Equation 3 by 2:
\[ 2t + 2b = 2 \cdot 3.75 = 7.50 \quad (Equation 4) \]
Now we have:
Next, we subtract Equation 4 from Equation 2:
\[ (4t + 2b) - (2t + 2b) = 10.00 - 7.50 \]
This simplifies to:
\[ 2t = 2.50 \]
Now, solving for \( t \):
\[ t = \frac{2.50}{2} = 1.25 \]
Now that we know \( t \), we can substitute back to find \( b \). Using Equation 3:
\[ 1.25 + b = 3.75 \]
Solving for \( b \):
\[ b = 3.75 - 1.25 = 2.50 \]
Thus, the costs are:
If you prefer the substitution method, you could solve Equation 3 for \( b \):
\[ b = 3.75 - t \]
Substituting \( b \) into Equation 2:
\[ 4t + 2(3.75 - t) = 10.00 \]
Expanding and simplifying:
\[ 4t + 7.50 - 2t = 10.00 \ 2t + 7.50 = 10.00 \ 2t = 10.00 - 7.50 \ 2t = 2.50 \ t = 1.25 \]
Now using \( t = 1.25 \) in \( b = 3.75 - t \):
\[ b = 3.75 - 1.25 = 2.50 \]
In either method, we find that the cost of one soft taco is \( t = 1.25 \) and the cost of one burrito is \( b = 2.50 \).
Certainly! Let's go through the elimination method step by step in detail for the problem involving the costs of soft tacos and burritos.
We started with our two equations:
Our goal is to eliminate one variable to solve for the other. We can do this by manipulating the equations in such a way that when we combine them, one of the variables cancels out.
First, we simplify Equation 1. Since all terms can be divided by 3, we do this:
\[ 3t + 3b = 11.25 \]
Dividing the entire equation by 3 gives:
\[ t + b = \frac{11.25}{3} = 3.75 \]
Let's call this Equation 3:
Now, we want to manipulate Equation 3 to facilitate the elimination of the variable \( b \).
To do this, we can multiply Equation 3 by 2 so that the coefficient in front of \( b \) matches that in Equation 2 when we subtract:
\[ 2(t + b) = 2(3.75) \]
This yields:
\[ 2t + 2b = 7.50 \quad \text{(Equation 4)} \]
Now we have:
Both equations have \( 2b \) on the left side. We are now ready to eliminate \( b \) by subtracting one equation from the other.
Now, we'll subtract Equation 4 from Equation 2:
\[ (4t + 2b) - (2t + 2b) = 10.00 - 7.50 \]
When we perform this subtraction, the \( 2b \) terms cancel out:
\[ 4t + 2b - 2t - 2b = 10.00 - 7.50 \]
This simplifies to:
\[ (4t - 2t) + (2b - 2b) = 2.50 \]
Which simplifies to:
\[ 2t = 2.50 \]
Now we can solve for \( t \):
\[ t = \frac{2.50}{2} = 1.25 \]
Now that we have \( t \), we can substitute it back into one of our earlier equations to find \( b \). We can use Equation 3:
\[ t + b = 3.75 \]
Substituting \( t = 1.25 \):
\[ 1.25 + b = 3.75 \]
Now solve for \( b \):
\[ b = 3.75 - 1.25 = 2.50 \]
We now have the costs as follows:
In summary, the elimination method involves:
This approach is efficient when the coefficients lend themselves to easy manipulation, as we showed in our example.
Absolutely! Let’s go through the substitution method step-by-step in detail for the problem involving the costs of soft tacos and burritos.
As a reminder, we’re working with the same two equations:
We start by solving one of the equations for one variable. We can simplify Equation 1 first to make things easier.
Simplifying Equation 1:
We can divide everything in Equation 1 by 3:
\[ 3t + 3b = 11.25 \]
Divide by 3:
\[ t + b = \frac{11.25}{3} = 3.75 \]
Let’s call this Equation 3 for reference:
Now, we can solve Equation 3 for \( b \):
\[ b = 3.75 - t \]
Next, we substitute what we found for \( b \) into Equation 2:
\[ 4t + 2b = 10.00 \]
Substituting \( b = 3.75 - t \):
\[ 4t + 2(3.75 - t) = 10.00 \]
Now, we simplify the equation:
\[ 4t + 7.50 - 2t = 10.00 \]
\[ 2t + 7.50 = 10.00 \]
\[ 2t = 10.00 - 7.50 \]
This simplifies to:
\[ 2t = 2.50 \]
\[ t = \frac{2.50}{2} = 1.25 \]
Now that we know the cost of one soft taco is \( t = 1.25 \), we can substitute this value back into our expression for \( b \) from Equation 3:
\[ b = 3.75 - t \]
Substituting \( t = 1.25 \):
\[ b = 3.75 - 1.25 \]
Thus:
\[ b = 2.50 \]
We have determined the costs:
In summary, the substitution method involves:
This method is particularly useful when one equation is already easily solvable for one of the variables, as we demonstrated in this example.
