What are the solutions to the system of equations?

Mathematics

1 answer:

bogdanovich [222]3 years ago

5 0

Answer:

(0, - 3 ) and (- 1, - 5 )

Step-by-step explanation:

Given the 2 equations

y = 4x² + 6x - 3 → (1)

y = 2x - 3 → (2)

Substitute y = 4x² + 6x - 3 into (2)

4x² + 6x - 3 = 2x - 3 ( subtract 2x - 3 from both sides )

4x² + 4x = 0 ← factor out 4x from each term

4x(x + 1) = 0

Equate each factor to zero and solve for x

4x = 0 ⇒ x = 0

x + 1 = 0 ⇒ x = - 1

Substitute these values into (2) for corresponding values of y

x = 0 : y = 2(0) - 3 = 0 - 3 = - 3 ⇒ (0, - 3 )

x = - 1 : y = 2(- 1) - 3 = - 2 - 3 = - 5 ⇒ (- 1, - 5 )

You might be interested in

If uq=6, find us and qs

Anon25 [30]

Can you provide more information?

4 0

4 years ago

on the last night of scout camp Chris his wife Joe and the 11 other adults sit down for supper along one side of a long rectangu

yawa3891 [41]

Based on the information provided, it follows that there are 1,728 possible seating arrangements.

<h3>How can we find the number of possible arrangements?</h3>

To find the number of arrangements in this problem situation we must take into account the following key factors:

Chris only has 1 possible seat.
Jo has 2 possible seats.
Dave, Alex, and Barb have 3 possible seats.
Gareth, Fred, and Eddie have 3 possible seats.
There are 4 other adults who do not have a preference in seats but have the possibility of using 4 seats.

According to the above, we must use the factorization of these numbers to find out the number of possibilities we have to seat them.

<h3>What is factoring?</h3>

A factorial function is a mathematical tool that is characterized by using the exclamation mark “!” behind a number. The factorial function is used to express that the number accompanied by the symbol must be multiplied by all positive integers between that number and 1.

In accordance with the above, in the problem situation that we must solve, we must use the factorial function with the possibilities of:

Dave, Alex and Barb: 3! = 3 × 2 × 1 = 6
Gareth, Fred and Eddie: 3! = 3 × 2 × 1 = 6
Other 4 adults: 4! = 4 × 3 × 2 × 1 = 24

Subsequently, to calculate the number of total possibilities of the entire group we must multiply the possibilities of each group and individual as shown below: