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kramer
3 years ago
13

Solve the following system. y = x 2 - 9x + 10 and x + y + 5 = 0. The solutions are

Mathematics
2 answers:
shepuryov [24]3 years ago
6 0
X+y+5=0

y=-x-5

If a solution exists y=y so we can say

x^2-9x+10=-x-5  add x+5 to both sides

x^2-8x+15=0 now factor

x^2-3x-5x+15=0

x(x-3)-5(x-3)

(x-5)(x-3) so x=3 and 5, using y=-x-5

y(3)=-8 and y(5)=-10

So the two solutions are:

(3,-8) and (5,-10)

ollegr [7]3 years ago
5 0

Answer:

Solution are ( 3,-8) (5 , -10).

Step-by-step explanation:

Given :  y = x² - 9x + 10 and x + y + 5 = 0.

To find : Solve the following system.

Solution : We have given that

y = x² - 9x + 10 -----( equation 1)

x + y + 5 = 0------( equation 2)

First we will solve equation 2 for y

on subtraction both sides by 5

x + y = -5

on subtracting both sides by x

y = -5 -x .

Now plug the value of y in equation 1

-5 -x = x² - 9x + 10

x² - 9x + x +5 + 10 = 0

x² - 8x + 15 = 0

Factoring  x² - 5x- 3x + 15 = 0

x ( x -5) -3 (x -5) =0

On grouping (x -3) (x -5)

For x = 3

y = -3 - 5

y = -8.

For x = 5

y = -5 -5

y = -10.

Solution are ( 3,-8) (5 ,- 10)

Therefore, Solution are ( 3,-8) (5 ,-10).

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The functions I(x) and E(x) of the theatre's income and expenses are illustrations of linear functions.

<h3>The equation of the theatre's income</h3>

The theatre charges $4.50 per student.

Assume the number of students is x, the equation of the theatre's income would be:

I(x) = 4.5x

<h3>The equation of the theatre's expenses</h3>

The theatre expense per student is $1.25, and the operating cost on the staff is $130

The equation of the theatre's expenses would be:

E(x) = 1.25x + 130

<h3>Complete the table</h3>

Using the formulas I(x) = 4.5x and E(x) = 1.25x + 130, the complete table is:

Students, x  0       10        20     30      40      50      60    70

Income, I      0      45        90     135     180    225    270   315

Expenses, E 130  142.5   155    167.5   180   192.5   205  217.5

<h3>The graph of the theatre's income and expenses</h3>

See attachment

<h3>The pattern by which theatre's income and expenses increase</h3>

The functions I(x) and E(x) are linear functions.

So, the pattern with which the functions increase is a linear pattern.

<h3>The number of students when the theatre's income and expenses are equal</h3>

This means that:

I(x) = E(x)

So, we have:

4.5x = 1.25x + 130

Subtract 1.25 from both sides

3.25x = 130

Divide both sides by 3.25

x = 40

Hence, the number of students is 40

<h3>The theatre profit</h3>

This is the difference between the theatre expenses and their income.

So, we have:

P(x) = E(x) - I(x)

This gives

P(x) = 1.25x + 130 - 4.5x

Simplify

P(x) = 130 - 3.25x

<h3>Solution to the inequalities</h3>

We have:

E(x) < 255

This gives

1.25x + 130 < 255

Subtract 130 from both sides and divide by 1.25

x < 100 students

Also, we have:

I(x) > 675

This gives

4.5x > 657

Solve for x

x > 146 students

Hence, the number of students for the inequalities are less than 100 and greater than 146

Read more about linear equations and inequalities at:

brainly.com/question/11234618

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