The product of t and s less than 3

Find the solution to the system of equations by using the substitution method<br> Y=4x+1<br> Y=x+1

Iteru [2.4K]

Answer:

x = 0

y = 1

Step-by-step explanation:

y = 4x + 1 -------eqn 1

y = x + 1 -------eqn 2

Looking for x, let's use eqn 1

y = 4x + 1

4x = y - 1

Divide both sides by 4, to get x

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

x = (y - 1) / 4

Substitute the value of x into eqn 2

y = x + 1

y = (y - 1) / 4 + 1

LCM = 4

y =( y - 1 + 4)/4

y =( y + 3) / 4

Cross multiply

y * 4 = y + 3

4y = y + 3

4y - y = 3

3y = 3

Divide both sides by 3, to get y

3y / 3 = 3/3

y = 1

Substitute y = 1 , into eqn 1

y = 4x + 1

1 = 4x + 1

1 -1 = 4x

0 = 4x

Divide both sides by 4, to get x

0/4 = 4x/4

0 = x

x = 0

Hint, let's check if the values are correct

Let's pick eqn 2

y = x + 1

y = 1

x = 0

1 = 0 + 1

1 = 1

Correct

Let's try the eqn 1

y = 4x + 1

1 = 4(0) + 1

1 = 0 + 1

1 = 1

Correct too

5 0

3 years ago

In a certain year, the U.S. Senate was made up of 53 Democrats, 45 Republicans, and 2 Independents who caucus with the Democrats

babunello [35]

Answer:

There is a 47.50% probability that the chosen senator is a Democrat.

Step-by-step explanation:

This can be formulated as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula:

$P = \frac{P(B).P(A/B)}{P(A)}$

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

In your problem we have that:

A(what happened) is the probability of a gun owner being chosen:

There are 100 people in the survay(53 Democrats, 45 Republicans ans 2 Independents), and 40 of them have guns(19 Democrats, 21 Republicans). So, the probability of a gun owner being chosen is:

$P(A) = \frac{40}{100} = 0.4$

$P(A/B)$ is the probability of a senator owning a gun, given that he is a Democrat. 19 of 53 Democrats own guns, so the probability of a democrat owning a gun is:

$P(A/B) = \frac{19}{53} = 0.3585$

$P(B)$ is the probability that the chosen senators is a Democrat. There are 100 total senators, 53 of which are Democrats, so:

$P(B) = \frac{53}{100} = 0.53$

If a senator participating in that survey was picked at random and turned out to be a gun owner, what was the probability that he or she was a Democrat?

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{(0.53)*(0.3585)}{(0.40)} = 0.4750$