Answer: Part A is 2 and 6 Part B is 2
Step-by-step explanation:
Part A: Here is the explanation. So, you started at with the expression 3x^2+8x+4 and when you're are factoring, you have 3x^2+px+pq+4. You can substitute the p and q for 6 and 2. What they did is they replaced 8x with px+qx. To get 8x, p needs to be 6 and q needs to be 2, or the other way around. TIP: The numbers just have to add up to 8 on this one. It doesn't have to be 6 and 2.
Part B: Here is what I got so far... 3x(x+r) is 3x^2+3xr. Also, s(x+r) is sx+sr. The equation becomes, 3x^2+3xr+sx+sr. R can be 2 and s can be 2. Here is my reasoning: The original expression was 3x^2+8x+4. We already have the 3x^2, so now we need to find what the others are by determining what r and s equal. R and s can both be 2 to make four. 2x2 is 4. Let's see if it can make 8. 3xr becomes 6x and sx becomes 2x. 6x+2x is 8x.
Answer:
watch the video
Step-by-step explanation:
Answer:
Both
Step-by-step explanation:
Solve the following system:{12 x = 54 - 6 y | (equation 1)-17 x = -6 y - 62 | (equation 2)
Express the system in standard form:{12 x + 6 y = 54 | (equation 1)-(17 x) + 6 y = -62 | (equation 2)
Swap equation 1 with equation 2:{-(17 x) + 6 y = -62 | (equation 1)12 x + 6 y = 54 | (equation 2)
Add 12/17 × (equation 1) to equation 2:{-(17 x) + 6 y = -62 | (equation 1)0 x+(174 y)/17 = 174/17 | (equation 2)
Multiply equation 2 by 17/174:{-(17 x) + 6 y = -62 | (equation 1)0 x+y = 1 | (equation 2)
Subtract 6 × (equation 2) from equation 1:{-(17 x)+0 y = -68 | (equation 1)0 x+y = 1 | (equation 2)
Divide equation 1 by -17:{x+0 y = 4 | (equation 1)0 x+y = 1 | (equation 2)
Collect results:Answer: {x = 4 {y = 1
Please note the { are supposed to span over both equations but it interfaces doesn't allow it. Please see attachment for clarification.
The first question is c i believe, I dont have time to answer the others