Respuesta :
We will use a system of equations to solve this. Let x be the number of basic phones sold and y be the number of upgraded phones sold. The equation to represent the sales would be
30x + 100y = 900 (because each basic phone sells for $30 and each upgraded phone sells for $100, and the store did $900 in sales that day).
The equation to represent the cost to the store would be
5x + 25y = 200 (because each basic phone costs the store $5 and each upgraded phone costs the store $25, and the store had a cost of $200 that day).
[tex] \left \{ {{30x+100y=900} \atop {5x+25y=200}} \right. [/tex]
We will use elimination to solve this. Make the coefficients of x the same by multiplying the bottom equation by 6:
[tex] \left \{ {{30x+100y=900} \atop {6(5x+25y=200}} \right. \\ \\ \left \{ {{30x+100y=900} \atop {30x+150y=1200}} \right. [/tex]
Now we can eliminate the x's by subtracting:
[tex] \left \{ {{30x+100y=900} \atop {-(30x+150y=1200)}} \right. \\-------- \\-50y=-300[/tex]
Divide both sides by 50:
[tex]\frac{50y}{50}= \frac{300}{50} y=6[/tex]
There were 6 upgraded phones sold on Tuesday.
30x + 100y = 900 (because each basic phone sells for $30 and each upgraded phone sells for $100, and the store did $900 in sales that day).
The equation to represent the cost to the store would be
5x + 25y = 200 (because each basic phone costs the store $5 and each upgraded phone costs the store $25, and the store had a cost of $200 that day).
[tex] \left \{ {{30x+100y=900} \atop {5x+25y=200}} \right. [/tex]
We will use elimination to solve this. Make the coefficients of x the same by multiplying the bottom equation by 6:
[tex] \left \{ {{30x+100y=900} \atop {6(5x+25y=200}} \right. \\ \\ \left \{ {{30x+100y=900} \atop {30x+150y=1200}} \right. [/tex]
Now we can eliminate the x's by subtracting:
[tex] \left \{ {{30x+100y=900} \atop {-(30x+150y=1200)}} \right. \\-------- \\-50y=-300[/tex]
Divide both sides by 50:
[tex]\frac{50y}{50}= \frac{300}{50} y=6[/tex]
There were 6 upgraded phones sold on Tuesday.
