Tea and Cookies
Tea costs $2 per cup and cookies cost $1 each. What equation represents spending exactly $6?
2t+1c=6
On a graph, x is cups of tea and y is cookies. What purchase does the point (1, 4) show?
That this person purchased 1 cup of tea and 4 cookies.
Divide both sides of 3(x - 4) = 45 by 3. What equation results?
x-4=15
Solve 21 = 3a .
a=7
Solve V = IR for R.
R=\frac{V}{I}
Tea costs $2 per cup and cookies cost $1 each. If you buy 1 cup of tea and spend $6 total, how many cookies do you buy?
4
For the graph of 2x + y = 6 , which point lies on it: (0, 3), (2, 2), or (3, 6)?
(2,2)
Distribute 3 in 3(x - 4) = 45 . What equation results?
3x - 12 = 45
Solve 7b - 4 = 10
b=2
Solve ax - c = d for x .
x=\frac{d+c}{a}
Tea costs $2 per cup and cookies cost $1 each. Does (2, 2), meaning 2 cups of tea and 2 cookies, cost $6? Explain your reasoning.
Yes.
2*2+2*1=4+2=6
For 2x + y = 6 , where x is tea and y is cookies, what does the y -intercept mean?
This means how many cookies you could get when getting no teas.
Starting with x - 4 = 15 , add 4 to each side. What equation results?
x = 19
Solve 3(x - 4) = 45
x=19
Solve ax - c = d for c.
c=ax-d
Tea costs $2 per cup and cookies cost $1 each. Do 3 cups of tea and 0 cookies cost exactly $6? Explain your reasoning.
Yes.
3*2+0*1=6
For 2x + y = 6 , where x is tea and y is cookies, what does the x -intercept mean?
This means how many teas you could get when getting no cookies.
A student claims 3(x - 4) = 45 becomes 3x - 4 = 45 . Explain the error.
They forgot to distribute the 3 to the -4.
Solve 5p + 8 = 3p + 20
p=6
Solve 3y + k = 12 for y.
y=\frac{12-k}{3}
Tea costs $2 per cup and cookies cost $1 each. List every whole-number purchase costing $6 as (tea,
cookies).
(3,0), (2, 2), (1, 4), (0,6)
On 2x + y = 6, x is whole cups of tea and y is whole cookies. Is (1.5, 3) a possible purchase? Explain.
Yes. 2*1.5+3=6
Is 6(x - 4) = 90 equivalent to 3(x - 4) = 45 ? Justify.
Yes, if you multiply both sides by 2 they are equivalent.
Solve 2(3m - 5) = 4m + 8
m=9
Solve 2(x + h) = q for h .
h=\frac{q}{2}-h