How to solve two-step equations - HSPT Math

This problem involves two seperate multiplication problems. Erin will make $450 at the reunion but supplies cost her $300 to make the shirts. So her profit is $150.

This is a least common multiple problem because we want to have the same number of each flower; meaning if I have 10 roses in each bouquet I should have 10 tulips and 10 daisies as well. In order to solve this problem we should break down the story problem. Let's look at the numbers we are having to work with: 6, 12, 18.

To do least common multiple we must look at the prime factors of each number and we can list them out. A factor is simply a number multiplied by a number to give us a product. A prime number is a number that contains only two factors, one of them being 1 and the other its own number.

So lets list the prime factors of 6, 12, and 18

(2 and 3 are both prime numbers, and factors of 6)

(2 x 2 = 4. 4x3=12 We have to say 2 x 2 because 4 is not a prime number).

Now we have the prime factors listed out for each of our numbers. Next is a fun trick. We must choose which number contains the most of each prime factor. In this case which number contains the most 2's? (12; because 12 has two 2 prime factors). Which number contains the most 3's? (18; because 18 has two 3 prime factors).

Our next step is to multiply the most of our prime factors so in this case:

36 is our least common multiple. So now what do you think we can do with this number? Well the 36 means that is the lowest number of flowers we need of each type in order to have an equal amount of each for the boquets.

Knowing this, if we need 36 roses, and we are able to buy 6 roses per bundle. We need 6 bundles of roses, because 36 divided by 6 is 6. If we get 12 tulips by the bundle, we take 36 divided by 12 to give us 3 bundles of tulips needed. Lastly we can buy 18 daisies per bundle, 36 divided by 18 gives us 2 bundles needed giving us our answers 6, 3, 2 (bundles of roses, tulips, and daisies).

, which rounds to

If there are 500 apples and 1 in 50 will remain unharvested, then we can find the number of unharvested apples by multiplying.

10 apples will remain unharvested. Of those, half will rot on the ground. Multiply to find how many apples rot on the ground.

5 apples will rot on the ground.

Since Gary will be 37 in five years, he is years old now. He is twice as old as Cathy, so she is years old, and in five years, she will be years old.

In ten years, Mark will be years old, so Mark is years old now, and Brian is one-third of this, or years old.

Let be the number of years in which Mark will be twice Brian's age. Then Brian will be , and Mark will be . Since Mark will be twice Brian's age, we can set up and solve the equation:

Mark will be twice Brian's age in years.

Let represent the unknown quantity.

The first expression:

"The product of a number and three" is three times this number, or

"Ten added to the product" is

The second expression:

"Twice the number" is two times the number, or

.

The desired equation is therefore

.

"The difference of a number and nine" is the result of a subtraction of the two, so we write this as

"Five-sevenths of" this difference is the product of and this, or

This is equal to forty, so write the equation as

Let represent the unknown number.

"The sum of a number and ten" is the expression . "Twice" this sum is two times this expression, or

.

"The difference of the number and one half" is a subtraction of the two, or

Set these equal, and the desired equation is

Let be the number of hours that Cliff parks his car in the garage. Since he pays a flat fee of plus per hour or portion thereof, he pays dollars. Since he does not want to spend more than ,

Since a portion of an hour counts as much as an hour, must be a whole number.

implies that

.

Cliff can park in the garage for a maximum of 6 hours, and since he entered the garage at , he must exit hours later:

,

or .

The -coordinate of the -intercept of the line of an equation can be found by setting and solving for :

The -intercept is the point , which is three units down from the origin. This is the point marked in the diagram.

When solving this problem, remember order of operations PEMDAS. The parentheses come first followed by the multiplication.

When solving this problem, remember order of operations PEMDAS. The parentheses come first followed by the multiplication, and then the addition.

When solving this problem, remember order of operations PEMDAS. The parentheses come first followed by the subtraction.

When solving this problem, remember order of operations PEMDAS. The parentheses come first followed by the multiplication.

When solving this problem, remember order of operations PEMDAS. The parentheses come first, followed by the multiplication, and then the division.

When solving this problem, remember order of operations PEMDAS. The parentheses come first followed by the multiplication.

When solving this problem, remember order of operations PEMDAS. The parentheses come first, followed by the division, and then the addition.