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Ksenya-84 [330]
1 year ago
9

Solve for x. 3x8 23 Choose 1 answer: B E x < OR 31 3 x < 5 31 5 ≤ x ≤- 3o| - 4x +266 - There are no solutions All values o

f a are solutions.​

Mathematics
1 answer:
jasenka [17]1 year ago
8 0

The solution of the inequality 3x - 8 <= 23 or -4x + 26 >= 6 is x <= 31/3

<h3>How to solve for x?</h3>

The inequality is given as:

3x - 8 <= 23 or -4x + 26 >= 6

Evaluate the like terms

3x <= 31 or -4x>= -20

Make x the subjects

x <= 31/3 or x <= 5

x <= 31/3 has a wider domain that x <= 5

Hence, the solution of the inequality 3x - 8 <= 23 or -4x + 26 >= 6 is x <= 31/3

Read more about inequality at

brainly.com/question/24372553

#SPJ1

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He rides 100 miles in 8 hours
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{4x-2y+5z=6 <br> {3x+3y+8z=4 <br> {x-5y-3z=5
lesya692 [45]

There are three possible outcomes that you may encounter when working with these system of equations:


  •    one solution
  •    no solution
  •    infinite solutions

We are going to try and find values of x, y, and z that will satisfy all three equations at the same time. The following are the equations:

  1. 4x-2y+5z = 6
  2. 3x+3y+8z = 4
  3. x-5y-3z = 5

We are going to use elimination(or addition) method

Step 1: Choose to eliminate any one of the variables from any pair of equations.

In this case it looks like if we multiply the third equation by 4 and  subtracting it from equation 1, it will be fairly simple to eliminate the x term from the first and third equation.

So multiplying Left Hand Side(L.H.S) and Right Hand Side(R.H.S) of 3rd equation with 4 gives us a new equation 4.:

4. 4x-20y-12z = 20      

Subtracting eq. 4 from Eq. 1:

(L.HS) : 4x-2y+5z-(4x-20y-12z) = 18y+17z

(R.H.S) : 20 - 6 = 14

5. 18y+17z=14

Step 2:  Eliminate the SAME variable chosen in step 2 from any other pair of equations, creating a system of two equations and 2 unknowns.

Similarly if we multiply 3rd equation with 3 and then subtract it from eq. 2 we get:

(L.HS) : 3x+3y+8z-(3x-15y-9z) = 18y+17z

(R.H.S) : 4 - 15 = -11

6. 18y+17z = -11

Step 3:  Solve the remaining system of equations 6 and 5 found in step 2 and 1.

Now if we try to solve equations 5 and 6 for the variables y and z. Subtracting eq 6 from eq. 5 we get:

(L.HS) : 18y+17z-(18y+17z) = 0

(R.HS) : 14-(-11) = 25

0 = 25

which is false, hence no solution exists



3 0
3 years ago
Subtract the equations.
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Answer:

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Step-by-step explanation:

Set them equal to each other, (5x+4y=25)-(5x+2y=3). Distribute the negative into the right equation, (5x+4y=25)(-5x-2y=-3). Then combine like terms to get your answer.

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TiliK225 [7]
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