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4 years ago

Solve the system by substitution. x – 3y = 4 2x – 6y = 8 A. no solution B. x = 4, y = 0 C. x = 0, y = 0 D. infinite solutions

Mathematics

2 answers:

Gwar [14]4 years ago

5 0

Answer:

Step-by-step explanation:

1. Isolate the x-term in the first equation:

x-3y=4 -> x=3y+4

2. Substitute 3y+4 in for x in the second equation and solve

2(3y+4)-6y=8 : Distribute the 2

6y+8-6y=8 : Combine like terms

8=8

3. Isolate the y-term in the first equation:

x-3y=4 -> -3y=-x+4

a. Solve for y:

(-3y=-x+4)/-3 : Divide by -3

y=(1/3)x-(4/3)

4. Substitute (1/3)x-(4/3) in for y in the second equation and solve:

2x-6((1/3)x-(4/3))=8 : Distribute the -6

2x-2x+8=8 : Combine like terms

8=8

Explanation:

When you solve a system of equations and get a true statement-

The lines are parallel.

Parallel lines will never intersect, so there is no solution.

ruslelena [56]4 years ago

4 0

Answer:

The correct answer is option D.

Step-by-step explanation:

if the pair of equation has one or more than one solution then it is said to be consistent.

Only one solution , then independent system.
More than one solution , then dependent system.

if the pair of equation has no solution then it is said to be inconsistent.

Given : x - 3y = 4 ...[1]

2x - 6y = 8 ...[2]

Solution :

Solving equations with the help of Substituting methods:

x - 3y = 4

x = 4 +3y

Putting value of x from [1] in [2]:

$2(4+3y)-6y=8$

$8+6y-6y=8$

$6y=6y$

0 = 0

Given , system of equation will have infinite solution. Hence consistent and dependent.

The given system of equations will have infinite solutions.

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Which of the following is an odd function

nika2105 [10]

Given the functions

(a) f(x) = x³ + 5x² + x

(b) f(x) = x² + x

Function (a)

f(-x) = (-x)³ + 5(-x)² + (-x)

= -x³ + 5x² - x

= -(x³ - 5x² + x)

The function is neither even nor odd.

Function (b)

f(-x) = (-x)² + (-x)

= -(-x² + x)

The function is neither even nor odd.

Function (c)

f(-x) = -(-x)

= x

= -f(x)

Because f(-x) = -f(x) the function is odd.

Answer: f(x) = -x is an odd function.

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3 years ago

Simplify 7.4z - 5 (-1.6z + 2.4) please

Sergeu [11.5K]

The answer is 15.4z-12

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3 years ago

PLS HELP MEEEEEEEEE pls i make brain

Degger [83]

Answer:

1) 0.75

2) y = $\frac{3}{4}$ x - 3

3) (4, 0); (0, -3)

Step-by-step explanation:

Convert to slope-intercept form (y = mx + b) by solving:

y = $\frac{3}{4}$ x - 3

m is the slope, so $\frac{3}{4}$ or you can write as 0.75.

To find the x and y intercepts, set the y value to zero and solve (x-intercept), and b is the y-intercept.

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3 years ago

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You and three of your friends go out to dinner during the break.the check comes to 131.85$ you all decide to leave a 20% tip for

almond37 [142]

Answer:

$105.48

Step-by-step explanation:

Multiply 131.85 by 0.2 to get 20% of 131.85.

You will get 26.37.

Since it is 4 people, multiply by 4.

You will get $105.48.

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3 years ago

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Suppose that five ones and four zeros are arranged around a circle. Between any two equal bits you insert a 0 and between any tw

PolarNik [594]

Answer:

Using backward reasoning we want to show that "We can never get nine 0's".

Step-by-step explanation:

Basically in order to create nine 0's, the previous step had to have all 0's or all 1's. There is no other way possible, because between any two equal bits you insert a 0.

If we consider two cases for the second-to-last step:

There were 9 0's:

We obtain nine 0's if all bits in the previous step were the same, thus all bit were 0's or all bits were 1's. If the previous step contained all 0's, then we have the same case as the current iteration step. Since initially the circle did not contain only 0's, the circle had to contain something else than only 0's at some point and thus there exists a point where the circle contained only 1's.

There were 9 1's:

A circle contains only 1's, if every pair of the consecutive nine digits is different. However this is impossible, because there are five 1's and four 0's (we have an odd number of bits!), thus if the 1's and 0's alternate, then we obtain that 1's that will be next to each other (which would result in a 1 in the next step). Thus, we obtained a contradiction and thus assumption that the circle contains nine 0's after iteratins the procedure is false. This then means that you can never get nine 0's.

To summarize, in order to create nine 0's, the previous step had to have all 0's or al 1's. As we didn't start the arrange with all 0's, the only way is having all 1's, but having all 1's will not be possible in our case since we have an odd number of bits.

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3 years ago