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5. Tierra, Nico, and Alex had [tex]$\$[/tex]865[tex]$ altogether. Tierra spent $[/tex]\frac{2}{5}[tex]$ of her money. Nico spent $[/tex]\[tex]$40$[/tex], and Alex spent twice as much as Tierra. If the three friends had the same amount of money left, how much money did Alex have in the beginning?

[tex]\[
\begin{array}{l}
T=\frac{3}{5}x \\
N=x-40
\end{array}
\][/tex]

Sagot :

GinnyAnsweridn
To determine how much money Alex had initially, let’s go through the problem step by step.

### Step 1: Define the Variables
Let [tex]\( T \)[/tex], [tex]\( N \)[/tex], and [tex]\( A \)[/tex] be the initial amounts of money Tierra, Nico, and Alex had, respectively. Let [tex]\( x \)[/tex] be the amount of money they each had left after spending.

### Step 2: Set Up the Equations
We know the total amount of money they had initially:
[tex]\[ T + N + A = 865 \][/tex]

We are given information about how much each person spent and how much they had left:
- Tierra spent [tex]\(\frac{2}{5}\)[/tex] of her money:
[tex]\[ T_{left} = T - \frac{2}{5}T = \frac{3}{5}T \][/tex]
- Nico spent [tex]\( \$40 \)[/tex]:
[tex]\[ N_{left} = N - 40 \][/tex]
- Alex spent twice as much as Tierra:
[tex]\[ A_{spent} = 2 \times \frac{2}{5}T = \frac{4}{5}T \][/tex]
So, Alex’s remaining money is:
[tex]\[ A_{left} = A - \frac{4}{5}T \][/tex]

Since after spending, they all had the same amount of money left:
[tex]\[ \frac{3}{5}T = N - 40 = A - \frac{4}{5}T \][/tex]

### Step 3: Solve the System of Equations
We now have three key equations:
[tex]\[ T + N + A = 865 \quad \text{(Equation 1)} \][/tex]
[tex]\[ \frac{3}{5}T = N - 40 \quad \text{(Equation 2)} \][/tex]
[tex]\[ \frac{3}{5}T = A - \frac{4}{5}T \quad \text{(Equation 3)} \][/tex]

#### Simplify Equation 3:
[tex]\[ \frac{3}{5}T + \frac{4}{5}T = A \][/tex]
[tex]\[ \frac{7}{5}T = A \][/tex]
[tex]\[ A = \frac{7}{5}T \quad \text{(Simplified form of Equation 3)} \][/tex]

#### Use Equation 2:
[tex]\[ N = \frac{3}{5}T + 40 \quad \text{(Rearranged form of Equation 2)} \][/tex]

#### Substitute [tex]\(N\)[/tex] and [tex]\(A\)[/tex] back into Equation 1:
[tex]\[ T + \left(\frac{3}{5}T + 40\right) + \frac{7}{5}T = 865 \][/tex]
[tex]\[ T + \frac{3}{5}T + 40 + \frac{7}{5}T = 865 \][/tex]
Combine like terms:
[tex]\[ T + \frac{3}{5}T + \frac{7}{5}T + 40 = 865 \][/tex]
[tex]\[ T + 2T + 40 = 865 \][/tex]
[tex]\[ 3T + 40 = 865 \][/tex]
Solve for [tex]\(T\)[/tex]:
[tex]\[ 3T = 825 \][/tex]
[tex]\[ T = 275 \][/tex]

Now use the value of [tex]\(T\)[/tex] to find [tex]\(N\)[/tex] and [tex]\(A\)[/tex]:
[tex]\[ N = \frac{3}{5} \times 275 + 40 \][/tex]
[tex]\[ N = 165 + 40 \][/tex]
[tex]\[ N = 205 \][/tex]

[tex]\[ A = \frac{7}{5} \times 275 \][/tex]
[tex]\[ A = 385 \][/tex]

### Conclusion
Therefore, Alex initially had:
[tex]\[ \boxed{385} \][/tex]

So, Alex had [tex]\( \$385 \)[/tex] in the beginning.
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