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3 years ago

3y = -x+11 2x+4y= 14 Solve the system.

Mathematics

2 answers:

Anna35 [415]3 years ago

5 0

Answer:

x = -1

y = 4

Step-by-step explanation:

3y = -x + 11

2x + 4y = 14

Rearrange equation 1

It becomes

x + 3y = 11

2x + 4y = 14

Multiply equation 1 by 4 and equation 2 by 3

We have

4x + 12y = 44

6x + 12y = 42

Subtract equation 2 from one

We have

-2x = 2

Divide both sides by-2

x = -2/2

x = -1

Now substitute-1 for x in any of the equations to get y .

Using equation 1, we have

-1 + 3y = 11

Add 1 to both sides of the equation

-1 + 1 + 3y = 11 + 1

3y = 12

Divide both sides by 3

3y/3 = 12/3

y = 4

trapecia [35]3 years ago

4 0

Answer:

x = -1 and y = 4

Step-by-step explanation:

Step 1: Solve 3y = -x + 11 for x

3y + x= -x + 11 + x

x + 3y = 11

x + 3y - 3y = 11 - 3y

x = −3y + 11

Step 2: Substitute -3y + 11 for x in 2x + 4y = 14

2(-3y + 11) + 4y = 14

Simplify both sides of the equation, and you get −2y + 22 = 14

−2y + 22 − 22 = 14 − 22

-2y = -8

-2y/-2 = -8/-2

y = 4

Step 3: Substitute 4 for y in x = −3y + 11

x = -3(4) + 11

Simplify both sides of the equation, and you get x = -1

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Please help!!As soon as possible!!

pychu [463]

Answer: Their equations have different y-intercept but the same slope

Step-by-step explanation:

Since, the slope of the line passes through the points (2002,p) and (2011,q) is,

$m_1 = \frac{q-p}{2011-2002} = \frac{q-p}{9}$

Similarly, the slope of the line asses through the points (2,p) and (11,q) is,

$m_2 = \frac{q-p}{11-2} = \frac{q-p}{9}$

Since, $m_1 = m_2$

Hence, both line have the same slope.

Now, the equation of the line one having slope $m_1$ and passes through the point (2002,p) is,

$y - p = \frac{q-p}{9}(x-2002)$

Put x = 0 in the above equation,

We get, $y = \frac{2000(p-q)}{9}+p$

The y-intercept of the line one is $(0, \frac{2000(p-q)}{9}+p)$

Also, the equation of second line having slope $m_2$ and passes through the point (2,p)

$y-p=\frac{q-p}{9}(x-2)$

Put x = 0 in the above equation,

We get, $y = \frac{2(p-q)}{9}+p$

The y-intercept of the line one is $(0, \frac{2(p-q)}{9}+p)$

Thus, both line have the different y-intercepts.

⇒ Third option is correct.

6 0

3 years ago

con 16 barras de mantequilla se prepararon 20 postres ?que fraccion del total de mantequilla corresponde a cada postre

MatroZZZ [7]

De ahí que la fracción de la mantequilla total que corresponde a cada postre sea 1/4

Si con 16 barras de mantequilla se prepararon 20 postres, esto se puede expresar como:

16 pegatinas = 20 postres

Para obtener la fracción del total de mantequilla que corresponde a cada postre, podemos escribir;

x = 1 postre

Divide ambas expresiones

16 / x = 20/1

20 veces = 16 * 1

20x = 16

x = 16/20

x = 4/5

De ahí que la fracción de la mantequilla total que corresponde a cada postre sea 1/4

Obtenga más información aquí: brainly.com/question/1878884

5 0

2 years ago

PLEASE HELP (30 POINTS) Solve the rational equation x/3 = x^2/x + 5 , and check for extraneous solutions.

anygoal [31]

Option C: $x=0$ and $x=\frac{5}{2}$ are the solutions.

Explanation:

The equation is $\frac{x}{3} =\frac{x^{2} }{x+5}$

We shall determine the value of x, by simplifying the equation.

$\begin{aligned} x(x+5) &=3 x^{2} \\ x^{2}+5 x &=3 x^{2} \\ 2 x^{2}-5 x &=0 \\ x(2 x-5) &=0 \end{aligned}$

Thus, $x=0$ and $x=\frac{5}{2}$ are the solutions.

Now, let us check whether the solutions are extraneous solutions.

Let us substitute $x=0$ in the original equation to check whether both sides of the equation are equal.

$\begin{aligned}&\frac{0}{3}=\frac{0^{2}}{0+5}\\&0=\frac{0}{5}\\&0=0\end{aligned}$

Thus, both sides of the equation are equal.

Hence $x=0$ is a true solution.

Now, Let us substitute $x=\frac{5}{2}$ in the original equation to check whether both sides of the equation are equal.

$\begin{aligned}\frac{\left(\frac{5}{2}\right)}{3} &=\frac{\left(\frac{5}{2}\right)^{2}}{\left(\frac{5}{2}\right)+5} \\\frac{5}{6} &=\frac{\left(\frac{25}{4}\right)}{\left(\frac{15}{2}\right)} \\\frac{5}{6} &=\frac{5}{6}\end{aligned}$

Thus, both sides of the equation are equal.

Hence, $x=\frac{5}{2}$ is a true solution.

Thus, solutions are not extraneous.

Hence, Option C is the correct answer.

8 0

3 years ago

Anyone know this I need help

horrorfan [7]

Answer:

option D

Step-by-step explanation:

we have a right triangle

we know

c^2=a^2+b^2

c=hypotenuse

c=52 cm

a=48 cm

b= 20 cm

we substitute in the equation c^2=a^a+b^2

so we have

52^2=48^2+20^2

52= sqrt(48^2+20^2)

52= sqrt(2304+400)

52= sqrt(2704)

52cm=52cm

so c=52 cm

so the option D is the correct answer

7 0

3 years ago

Read 2 more answers

How many times longer is one centimeter than one millimeter

Nastasia [14]

A kilogram is 1,000 times larger than one gram (so 1 kilogram = 1,000 grams). A centimeter is 100 times smaller than one meter (so 1 meter = 100 centimeters). A dekaliter is 10 times larger than one liter (so 1 dekaliter = 10 liters).

3 0

3 years ago