Respuesta :
Answer:
Step-by-step explanation:
Let:
i: the cost of one pot of ivy.
r: the cost of one rose bush
[tex]5r+3i=58 ......(1)\\5r+10i=100.......(2)\\\\[/tex]
From eq(1) we can get:
[tex]r=\frac{58-3i}{5} \\[/tex]
Substitute eq(1) in eq(2):
[tex]5(\frac{58-3i}{5})+10i=100\\58-3i+10i=100\\7i=100-58\\i=\frac{42}{7} =6\\[/tex]
[tex]r=\frac{58-3i}{5} \\r=\frac{58-3(6)}{5} \\\\r=8[/tex]
Answer: A rose bush, r, cost $8; one pot of ivy, i, cost $6;
Step-by-step explanation:
[tex]\left \{ {{5r + 3i =58} \atop {5r+10i=100}} \right.[/tex]
We combine them by minus, we get:
5r + 3i = 58
- 5r + 10i = 100
- 7i = - 42
i = 6
We plug in i = 6 to the first equation, we get:
5r + 3(6) = 58
5r + 18 = 58
5r = 40
r = 8