A)
s = sweater
j = jeans
Equations:
4s + 2j = 140
2s + 3j = 150
B)
4s + 2j = 140
2s + 3j = 150
Multiply (-2) to the second equation
-2 (2s + 3j = 150)
-4s - 6j = -300
Now you have the equations:
4s + 2j = 140
-4s - 6j = -300
-----------------------------add
-4j = -160
j = 40
4s + 2(40) = 140
4s + 80 = 140
4s = 60
s = 15
The cost for a sweater is $15 and the cost for a jean is $40
We used the elimination method to solve the system of equations.
C)
Each pair of jean, she will buy 3 sweaters so total = 40 + 3(15) = $85
Total cost for 1 pair of jeans and 3 sweaters = $85
If she has $225 the
$225 / $85 = 2.64
2 jeans = 2 x 40 = $80
6 sweaters = 6 x $15 = $90
So She can only buy 2 jeans and 6 sweaters
The answer would be 6b-3c because a would cancel out and be 0
You can rewrite the multiplication using properties of the exponents.
We have that if the base is the same, the exponents are added.
For this case, we have:
Base = (- 4)
Exponent = 1
Applying properties of exponents:
(-4) (- 4) = ((- 4) ^ 1) ((- 4) ^ 1) = (- 4) ^ (1 + 1) = (- 4) ^ 2
answer:
the expression is the same as (-4) ^ 2
