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3 years ago

Solve for x 3 x − 8 ≤ 23 AND − 4 x + 26 ≥ 6 Please help thank you!!

Mathematics

1 answer:

MAXImum [283]3 years ago

8 0

To solve for x, you need to isolate/get the variable "x" by itself in the inequality:

3x - 8 ≤ 23 Add 8 on both sides

3x - 8 + 8 ≤ 23 + 8

3x ≤ 31 Divide 3 on both sides to get "x" by itself

$\frac{3x}{3} \leq \frac{31}{3}$

$x\leq \frac{31}{3}$ (x is any number less than or equal to 31/3)

-4x + 26 ≥ 6 Subtract 26 on both sides

-4x ≥ -20 Divide -4 on both sides [When you multiply/divide by a negative number to both sides of an inequality, you flip the sign </>]

$\frac{-4x}{-4} \geq \frac{-20}{-4}$ (two negative signs cancel each other out and become positive)

x ≤ 5 (x is any number less than or equal to 5)

You might be interested in

What is 26 divided by 10.92

Bezzdna [24]

26 divided by 10.92 equals 2.380952

7 0

3 years ago

PLEASE HELP WITH GRADE 11 MATH. Make sure you show the formula. Substitute values and show all mathematicall operations! show yo

Degger [83]

3x+y

=−3

=−y+3

The second equation is solved for xxx, so we can substitute the expression -y+3−y+3minus, y, plus, 3 in for xxx in the first equation:

\begin{aligned} 3\blueD{x}+y &= -3\\\\ 3(\blueD{-y+3})+y&=-3\\\\ -3y+9+y&=-3\\\\ -2y&=-12\\\\ y&=6 \end{aligned}

3x+y

3(−y+3)+y

−3y+9+y

−2y

=−3

=−12

Plugging this value back into one of our original equations, say x = -y +3x=−y+3x, equals, minus, y, plus, 3, we solve for the other variable:

\begin{aligned} x &= -\blueD{y} +3\\\\ x&=-(\blueD{6})+3\\\\ x&=-3 \end{aligned}

=−y+3

=−(6)+3

=−3

The solution to the system of equations is x=-3x=−3x, equals, minus, 3, y=6y=6y, equals, 6.

We can check our work by plugging these numbers back into the original equations. Let's try 3x+y = -33x+y=−33, x, plus, y, equals, minus, 3.

\begin{aligned} 3x+y &= -3\\\\ 3(-3)+6&\stackrel ?=-3\\\\ -9+6&\stackrel ?=-3\\\\ -3&=-3 \end{aligned}

3x+y

3(−3)+6

−9+6

−3

=−3

−3

=−3

Yes, our solution checks out.

Example 2

We're asked to solve this system of equations:

\begin{aligned} 7x+10y &= 36\\\\ -2x+y&=9 \end{aligned}

7x+10y

−2x+y

=36

In order to use the substitution method, we'll need to solve for either xxx or yyy in one of the equations. Let's solve for yyy in the second equation:

\begin{aligned} -2x+y&=9 \\\\ y&=2x+9 \end{aligned}

−2x+y

=2x+9

Now we can substitute the expression 2x+92x+92, x, plus, 9 in for yyy in the first equation of our system:

\begin{aligned} 7x+10\blueD{y} &= 36\\\\ 7x+10\blueD{(2x+9)}&=36\\\\ 7x+20x+90&=36\\\\ 27x+90&=36\\\\ 3x+10&=4\\\\ 3x&=-6\\\\ x&=-2 \end{aligned}

7x+10y

7x+10(2x+9)

7x+20x+90

27x+90

3x+10

=36

=−6

=−2

Plugging this value back into one of our original equations, say y=2x+9y=2x+9y, equals, 2, x, plus, 9, we solve for the other variable:

\begin{aligned} y&=2\blueD{x}+9\\\\ y&=2\blueD{(-2)}+9\\\\ y&=-4+9 \\\\ y&=5 \end{aligned}

=2x+9

=2(−2)+9

=−4+9

The solution to the system of equations is x=-2x=−2x, equals, minus, 2, y=5y=5y, equals, 5.

5 0

2 years ago

What value is the cube root of 27

umka21 [38]

3 is the cube root of 27

7 0

3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=%20%7Bx%7D%5E%7B2%7D%20-%2022x%20%20%2B%20121" id="TexFormula1" title=" {x}^{2} - 22x + 121"

eduard

Let's factor x^2−22x+121

The middle number is -22 and the last number is 121. Factoring means we want something like

(x+_)(x+_)

Which numbers go in the blanks? We need two numbers that...

Add together to get -22

Multiply together to get 121

Can you think of the two numbers?

Try -11 and -11:

-11+-11 = -22

-11*-11 = 121

Fill in the blanks in

(x+_)(x+_)

with -11 and -11 to get...

(x-11)(x-11)

Answer: (x-11)(x-11)

Hope this helps!!!

3 0

3 years ago

A jar contains 8 red marbles numbered 1 to 8 and 7 blue marbles numbered 1 to 7. A marble is drawn at random from the jar. Find

Shkiper50 [21]

Answer:

Therefore the probability that the marble is blue or even numbered is $\frac{11}{15}$

Step-by-step explanation:

Probability: The ratio of favorable outcomes to the total outcomes.

It is denoted by P.

$Probability= \frac{\textrm{favorable outcomes}}{\textrm{Total outcomes}}$

Given that a jar contains 8 red marbles and 7 blue marbles.

Total number of marbles = (8+7) = 15

Let A = Event of getting a blue marble

B= Event of getting of even marble.

Even number blue marbles are 2, 4,6

Even number red marbles are 2, 4,6,8

The number of even marbles are =(3+4)=7

The probability of getting a blue marble is P(A)

$=\frac{\textrm{Total number of blue marbles}}{\textrm{Total number of blue marbles}}$

$=\frac{7}{15}$

The probability of getting a even marble is P(B)

$=\frac{\textrm{The number of even number marbles}}{\textrm{Total number of marbles}}$

$=\frac{7}{15}$

The probability of getting a even numbered blue marble P(A∩B)

$=\frac{3}{16}$

P(blue marble or even- numbered)

=P(A∪B)

=P(A)+P(B)-P(A∩B)

$=\frac{7}{15} +\frac{7}{15}-\frac{3}{15}$

$=\frac{11}{15}$

Therefore the probability that the marble is blue or even numbered is $\frac{11}{15}$

3 0

3 years ago