Question

How many solutions does the system have?
y = -3x+9.
3y=-9x+9​

Answer 1

Let's bring both equations to slope-intercept form. Then we can think about the slopes and the yyy-intercepts of the lines represented by each equation.

The first equation y = -3x+9y=−3x+9y, equals, minus, 3, x, plus, 9 is already in slope-intercept form. The slope-intercept form of the second equation 3y=-9x+93y=−9x+93, y, equals, minus, 9, x, plus, 9 is y=-3x+3y=−3x+3y, equals, minus, 3, x, plus, 3.

Hint #22 / 3

The first equation is y = -3x+9y=−3x+9y, equals, minus, 3, x, plus, 9, so the slope of its line is -3−3minus, 3 and the yyy-intercept is (0,9)(0,9)left parenthesis, 0, comma, 9, right parenthesis.

The second equation is y = -3x+3y=−3x+3y, equals, minus, 3, x, plus, 3, so the slope of its line is -3−3minus, 3 and the yyy-intercept is (0,3)(0,3)left parenthesis, 0, comma, 3, right parenthesis.

Since both lines have the same slopes but different yyy-intercepts, they are distinct parallel lines.

Hint #33 / 3

Answer:

Since distinct parallel lines don't intersect, we conclude that the system has no solutions.

From Khan Academy

Answer 2

No solutions

Step-by-step explanation:

`\left \{ {{y=-3x+9} \atop {3y=-9x+9}`

1. Divide the second equation by 3

3y=-9x+9​

y=-3x+3

`\left \{ {{y=-3x+9} \atop {y=-3x+3}`

2. Plug the first equation into the second as y

`-3x+9=-3x+3`

-3 cancels out on both sides (since it's on both sides of the equation)

`9=3`

`9\\eq 3`

Because the left side of the equation doesn't equal the right side, we can conclude that the system of equations has no solution.

I hope this helps! Please comment if you have any questions :)