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April 1, 2015

April 1, 2015

Posted by **Bri** on Monday, January 31, 2011 at 4:52pm.

- algebra -
**helper**, Monday, January 31, 2011 at 5:12pmb = number of blooming annuals

n = number of non-blooming annuals

3.20b = value of blooming

1.50n = value of non-blooming

b + n = 24

3.20b + 1.50n = 49.60

Solve the system

- algebra -
**Bri**, Monday, January 31, 2011 at 5:49pmi do not think that is what my teacher was asking for.

- algebra -
**helper**, Monday, January 31, 2011 at 5:56pm"write a linear system of equations"

This is a linear system of equations.

You can use x and y instead of n and b.

x = number of blooming annuals

y = number of non-blooming annuals

3.20x = value of blooming

1.50y = value of non-blooming

x + y = 24

3.20x + 1.50y = 49.60

Then solve the system.

Have you learned solving simultaneous equations?

This can also be solved using one equation, but the problem asked for a linear system of equations.

- algebra -
**Bri**, Monday, January 31, 2011 at 6:04pmya i did today. but i dont understand how you got 24 as the y-intercept

- algebra -
**helper**, Monday, January 31, 2011 at 6:25pm24 is the total number of plants.

x = number of blooming annuals

y = number of non-blooming annuals

x + y = 24

- algebra -
**Bri**, Monday, January 31, 2011 at 6:27pmokaaaay im starting to get it. so what would the second equation be again? and can you please explain it? =) thanks!

- algebra -
**helper**, Monday, January 31, 2011 at 8:00pm3.20x = value of blooming annuals

3.20 each plant * x number of plants = ttl cost

1.50y = value of non-blooming annuals

1.50 each plant * y number of plants = ttl cost

3.20x + 1.50y = 49.60

- algebra -
**Bri**, Monday, January 31, 2011 at 9:07pmok thanks!

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