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>
It is called an equation.
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You got to shade five squares
Answer:
Step 2 contains error in the given problem.
Step-by-step explanation:
Given expression is:
Step 1: identifying the LCM.
The LCM identified is 6.
This step is correct.
In the next step, we multiply the LCM with each term of the equation.
Step 2:
However,
In the given solution, the LCM is not multiplied with each term.
Hence,
Step 2 contains error in the given problem.
