For an end-of-year party, Ms. Carter purchased 98 stickers and 56 lollipops. Each student will receive an equal number of stickers and an equal number of lollipops. There will be no stickers or lollipops leftover. Based on this information, what is the greatest number of students that Ms. Carter can have in her class?

Example 1: One-Step Inequalities

A bag of candy is split between us and our little brother. The bag says it has at most 28 pieces of candy in it. There are 15 candies in our bag, and x candies in our brother's bag, and we have to make sure he doesn't have more than us (of course). We can write an inequality expressing the number of candies in the bags.

We know that our bag has 15 candies.

We know that his bag has x candies.

We know that up to 28 candies were split between both bags, no more.

Since we know that x + 15 can't be more than 28, it must be less than or equal to 28. x + 15 ≤ 28

Now that we have an inequality expressing our candy, we solve for x. To do that, we subtract 15 from both sides.

x + 15 ≤ 28

x + 15 – 15 ≤ 28 – 15

x ≤ 13

Sweet, literally. Our brother has 13 or fewer pieces of candy, so no way does he have more candy than us. Until, in a moment of weakness (er, kindness) we share a few of our pieces with him. Now it's even sweeter.

Example 2: Two-Step Inequalities

A candy store owner saw us share our candy with our brother and was so impressed he gave us a $30.00 gift card. Tax free even! We decide to buy a giant candy bar for $13.00 and then some lollipops with the remaining money. If each lollipop is $0.90, we can write and solve an inequality expressing how many lollipops we can buy.

We know that the amount we spend needs to be less than or equal to $30.00.

We know we want to buy x lollipops.

We know that each lollipop is $0.90, so the total cost of the lollipops is $0.90x.

We know the candy bar costs $13.00.

So the total cost of the candy is $13 + $0.90x, and this needs to be less than or equal to $30.00. 13 + 0.90x ≤ 30

Now we solve for x, which is the number of lollipops we can buy. We want to get x by itself, so start by subtracting 13 from both sides.

13 + 0.90x ≤ 30

13 + 0.90x – 13 ≤ 30 – 13

0.9x ≤ 17

Divide both sides by 0.9 to finish up. We recommend nabbing a calculator for this part.

We need to buy less than 18.89 lollipops, so we can buy 18 whole lollipops and have some change to spare.

5x + 5

step-by-step explanation:

you should always add the diagrams and graphs we cannot see. this particular question is kind of borderline because we can make a graph for you.

(f + g)x = -x^2 +3x + 5 + x^2 + 2x

(f + g)x = 5x + 5

the graph of z = 5x + 5 is shown below. always use desmos when in doubt.

a.   we have two

lines:   y = 4-x   and  

y = 8-x^-1

given two simultaneous equations that are both to be

true, then the solution is the points where the lines cross. the intersection is

where the two equations are equal. therefore the solution that works for both

equations is when

4-x = 8-x^-1

this is where the two lines will cross and that is the

common point that satisfies both equations.

b.   4-x = 8-x^-1

  x           4-x       8-x^-1

-3           7        

8.33

-2           6         8.5

-1           5         9

  0           4        

-

  1           3        

7

  2           2        

7.5

  3           1         7.67

the table shows that none of the x values from -3 to 3 is

the solution because in no case does

4-x = 8-x^-1

to find the solution we need to rearrange the equation to

find for x:

4-x = 8-x^-1

multiply both sides with x:

4x-x^2 = 8x-1

x^2+4x-1=0

x= -4.236, 0.236

therefore there are two points that satisfies the

equation.

find y:  

x=-4.236

y = 4-x   = 4 – (-4.236)

= 8.236

y = 8-x^-1 =   .236)^-1

= 8.236

x=0.236

  y = 4-x   = 4 – (0.236) = 3.764

y = 8-x^-1 =   8-(0.236)^-1

= 3.764

thus the two lines cross at 2 points:

(-4.236, 8.236) & (0.236, 3.764)

c.   to solve

graphically the equation 4-x = 8-x^-1

we would graph both lines: y = 4-x   and   y

= 8-x^-1

the point on the graph where the lines cross is the

solution to the system of equations.

just graph the points on part b on a cartesian coordinate

system and extend the two lines.   the

solution is, as stated, the point where the two lines cross on the graph.

step-by-step explanation:

mark brainliest!