Question

I just need help with three, four, and seven. Solve the system of linear equations, and determine if the equation has one solution, infinitely many solutions, or no solution. Please, explain how you’ve gotten your answer. Thank you so much to whoever takes the time to help me out!

Answer 1

To find the equations that have no solution, solve each equation.

Solve the equation from question 1, part B:

5y − 1 = 5y + 1

5y − 5y = 1 + 1

0 = 2

Because 0 can never be equal to 2, the equation can never be true for any value of y. So, this equation has no solution.

Solve the equation from question 1, part C:

5y − 1 = 4y + 5

5y − 4y = 5 + 1

y = 6

This is a value of y for which the equation holds true, so this equation has one unique solution.

Solve the equation from question 1, part D:

5y + 1 = 4y + 5

5y − 4y = 5 − 1

y = 4

This is a value of y for which the equation holds true, so this equation has one unique solution.

Therefore, only one equation has no solution: 5y − 1 = 5y + 1.

Explanation:

for the people on edmentum heres the sample might want to change it around so you wont get caught

Answer 2

3. If
`y=5x+9` and
`y=3x-2`, then

`5x+9=3x-2\implies2x+9=-2\implies2x=-11\implies x=-\frac{11}2`

Then

`y=5\left(-\frac{11}2\right)+9=-\frac{55}2+9=-\frac{37}2`

(only one solution)

4. If
`y+2=4x`, then
`y=4x-2`. So we're told that
`4x-5=4x-2` but this is impossible, because that would that for any
`x`, we have -5 = -2. This is false, so this system has no solution.

7. "y is 6 less than 3 times x" means
`y=3x-6`. "x is 2 more than one-third of y" means
`x=\frac13y+2`. Solving the second equation for
`y` gives
`y=3x-2`, which would mean
`3x-6=3x-2`. But for the same reason as in (4), there is no solution. For any value of
`x`, this equation gives the false conclusion that -6 = -2.