7*11=77 because 7 and 11 are 4 apart Enjoy!=)
Sole for f by simplifying both sides of the equation then isolating the variable so that means f=-40 over 3 for number one
for the second equation you do the same thing you did for number one and the answer will be
b=8
March = -165
April = -156
In march he is negative MORE money, meaning he has MORE money in April
This question is trying to confuse you, ussually the lowest number means less, but these numbers are negative, so the lowest number means more.
He has the least amount of money in March
Answer:
I would multiply the first equation by −4 and the second by 3 and add together the two equatins (in columns): ... Step 2. Prepare the equations. Multiply every term in each equation by a ... Subtract Equation (4) from Equation (3). ... How do you solve the system 5x−10y=15 and 3x−2y=3 by multiplication?
Answer:
Let's define the cost of the cheaper game as X, and the cost of the pricer game as Y.
The total cost of both games is:
X + Y
We know that both games cost just above AED 80
Then:
X + Y > AED 80
From this, we want to prove that at least one of the games costed more than AED 40.
Now let's play with the possible prices of X, there are two possible cases:
X is larger than AED 40
X is equal to or smaller than AED 40.
If X is more than AED 40, then we have a game that costed more than AED 40.
If X is less than or equal to AED 40, then:
X ≥ AED 40
Now let's take the maximum value of X in this scenario, this is:
X = AED 40
Replacing this in the first inequality, we get:
X + Y > AED 80
Replacing the value of X we get:
AED 40 + Y > AED 80
Y > AED 80 - AED 40
Y > AED 40
So when X is equal or smaller than AED 40, the value of Y is larger than AED 40.
So we proven that in all the possible cases, at least one of the two games costs more than AED 40.
