8 to 16
2 to 4
13 to 26
5 to 10
6 to 12
A. Alright, we want to multiply one equation by a constant to make it cancel out with the second. Since the first equation has a "blank" y, let's multiply the first equation by <em>2</em>.
3x-y=0 → 2(3x-y=0) = 6x - 2y = 0
5x+2y=22
The answer for this part would be: 6x - 2y = 0 and 5x + 2y = 22
B. So now we combine them:
6x - 2y = 0
+ + +
5x + 2y = 22
= = =
11x + 0 = 22 ← The answer
C. Now that we have the equation 11x = 22, we solve for x
11x = 22 ← Divide both sides by 11
x = 2 ← The answer
D. Now that we have x=2, we plug that back in to 5x+2y=22 and solve for y:
5(2)+2y = 22
10 + 2y = 22
2y = 12
y = 6
<u>Therefore, the solution to this problem is x = 2 and y = 6</u>
Answer:
possibly 5?
Step-by-step explanation:
But can you plzzz help me with my latest question
Answer:
The coordinates of the image of point A (2, -7) are A'(-1,-2).
Step-by-step explanation:
Note: The sign is missing between y and 5 in the rule of transitional.
Consider the rule of translation is
We need to find the image of point A (2, -7).
Substitute x=2 and y=-7 in the above rule.
Therefore, the coordinates of the image of point A (2, -7) are A'(-1,-2).
