All categories

1 year ago

Solve the system of equations: y= x^2 + 2x - 3 and y=3x+ 3.

Mathematics

1 answer:

Semenov [28]1 year ago

3 0

Answer:

(-2, -3)

(3, 12)

Step-by-step explanation:

To solve this, we're gonna get rid of the y's with substitution

x² + 2x - 3 = 3x + 3

Let's make this equation equal to zero

Subtract 3 from both sides

x² + 2x - 3 = 3x + 3

- 3 - 3

x² + 2x - 6 = 3x

Subtract 3x from both sides

x² + 2x - 6 = 3x

- 3x - 3x

x² - x - 6 = 0

Factor the equation

(x - 3)(x + 2) = 0

This means x can be -2 or 3

Let's solve it with -2 first, plug the new x in y = 3x + 3

y = 3(-2) + 3

y = -6 + 3 = -3

Do the same for x = 3

y = 3(3) + 3

y = 9 + 3 = 12

You might be interested in

HELP DUE IN 5 MINUTES

iris [78.8K]

Answer:

4cm

Step-by-step explanation:

The base of the triangle is 4cm. 14-10 = 4cm

5 0

2 years ago

Read 2 more answers

Make x the subject of the formula y= 2x +5

Kryger [21]

Answer:

x= (y-5)/2

Step-by-step explanation:

Isolate the x variable.

y= 2x +5

Subtract 5 from both sides.

y-5=2x

Divide both sides by 2.

(y-5)/2= x

5 0

1 year ago

Katen ran 5 miles. How far did She run in kilometers

Assoli18 [71]

8.04672 hope this helps :D

3 0

2 years ago

Read 2 more answers

Which of the following equations is equivalent to 6.87p + 9.2 = -7.056?

Lina20 [59]

Answer:

The answer is the second one 687p + 920 = -705.6

Step-by-step explanation:

The decimal place in each number was shifted right 2 spots

3 0

2 years ago

A company manufactures and sells x television sets per month. The monthly cost and price-demand equations are C(x)=72,000+60x

myrzilka [38]

Answer:

Step-by-step explanation:

Given

Cost Price $c(x)=72000+60x$

Price $p(x)=300-\frac{x}{20}$

Revenue generated $R(x)=P(x)\times x$

where x=no of units

$R(x)=300x-\frac{x^2}{20}$

To get maxima and minima differentiate R(x)

$\frac{\mathrm{d} R(x)}{\mathrm{d} x}=0$

$\frac{\mathrm{d} R(x)}{\mathrm{d} x}=300-2\times \frac{x}{20}=0$

$300=2\times \frac{x}{20}$

$x=3000$

maximum Revenue $R(x)=(300-\frac{300}{20})\times 300=4,50,000$

(b)Profit=Revenue - cost

$Profit=xp(x)-c(x)$

$Profit=300x-\frac{x^2}{20}-72000-60x$

$Profit(z)=240x-\frac{x^2}{20}-72000$

differentiate Profit to get maximum value

$\frac{\mathrm{d} z}{\mathrm{d} x}=240-2\times \frac{x}{20}$

$x=2400$

maximum Profit $z=2,16,000$

(c)Now company decided to tax the company $ 55 for each set

Profit $(z_1)=xp(x)-c(x)-55x$

$z_1=300x-\frac{x^2}{20}-72000x-60x^2-55x$

$z_1=185x-\frac{x^2}{20}-72,000$

differentiate Profit to get maximum value

$\frac{\mathrm{d} z_1}{\mathrm{d} x}=0$

$\frac{\mathrm{d} z_1}{\mathrm{d} x}=185-\frac{2x}{20}=0$

$x=1850$

$P(z_1\ at\ x=1850)=99125$

company should charge 207.5 $ for each set

6 0

2 years ago