Answer:
Explanation:
Q1) We know that y = 2x+8, and y = 2x+2, this means that the equations should be equivalent (they both = y)
2x + 8 = 2x + 2
This is impossible, so there are no solutions. (Try plugging in for x if you don't get it - answering fast as per your request!)
Q2)
We can rearrange the first equation. -x - y = 1
1. Add y to both sides
2. Subtract 1 from both side
So now we have : y = -x-1
y = x + 3
These intersect when again, they are equivalent so we solve the equation:
x + 3 = -x-1
2x + 3 = -1
2x = -4
x = -2
So the answer must be (1,-2) ... (plug x back in for y usually to get the points, but here it's MC and only one has x = -2)
Q3)
2x + 2y = 8 - Line A can be divided by 2 to look more like Line B
Line A = x+y = 4
Similar to problem 1. x+y cannot equal both 3 AND 4, there is no solution.
Q4)
Again, same concept as problem 1. Both A and B are equal to Y, so we can find the solution by setting the equal:
-x +5 = 6x -2
Q5)
Same thing!
-x +2 = 3x +1
4x + 1 = 2
4x = 1
x = 1/4
This means that the two lines must intersect at some point, the point at which two lines intersect is the solution to their systems.
Line y = −x + 2 intersects line y = 3x + 1.
Q6)
Q = 0.5x + 3
S = 0.5x - 2
Lines Q and S have the same slope but different y-intercepts. This means they are parallel and will never intersect, so they are no solutions for their system of equations.
Q7)
Substitution means we want to solve for a variable in one equation, and plug this into the second, so we obtain a solvable, 1 variable equation.
We know y = 3x +3, and our second equation is equal to y. So we can substitute this y for 3x +3.
EQ1: y = 3x +3
EQ2: y = x-1 (substituting y for 3x+3 into this equation)
3x +3 = x - 1
-x -x
-3 -3
2x = -2
x = -1
plugging this into the simpler equation:
y = (-1) -1
y = -2
So the solution is (-1,-2).
Hope I answered it in time and you can make up an excuse if it's a little late!