A student is trying to solve the system of two equations given below: Equation P: y + z = 6 Equation Q: 3y + 4z = 1 Which of the
2 answers:
Answer:
Option C is correct
(y + z = 6) ⋅ −3
Step-by-step explanation:
As per the statement:
A student is trying to solve the system of two equations given below:
Equation P: y + z = 6 ....[1]
Equation Q: 3y + 4z = 1 ....[2]
Multiply the equation [1] by -3 to both sides we have;
or
......[3]
Add equation [2] and [3] to eliminate the y-term;
or
z = -17
Therefore, the possible step used in eliminating the y-term is, (y + z = 6) ⋅ −3
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- Put brackets around the first two terms on the right. y = (-4x^2 - 16x) - 14
- Pull out the common factor in the first two terms. y = -4(x^2 + 4x) - 14
- Divide the middle term by 2. Add that inside the brackets. square y = -4(x^2 +4x +(4/2)^2 - 14
- y = -4(x^2 + 4x + 4) - 14 + 16 This is the step where most people stumble. The point is why is 16 added? It is because what you have done inside the brackets is multiplied 4 by - 4 (on the left). That means you have changed the equation by -16. To counter that, you must add 16 after the - 14. The result is y= -4(x^2 + 4x+4) + 2
- Express the terms inside the brackets as a square. y = - 4(x + 2)^2 + 2
B and D are both wrong. B has +4 outside the brackets. It is - 4
D is wrong because there is no 4 of any kind outside the brackets.
A is incorrectly represented inside the brackets as 16. That's not right
That only leaves C. <<<< Answer
Graphs
Notice that the red parabola and the green one are the same thing.
Red: y = -4x^2 - 16x - 14
Green: y = -4(x + 2)^2 + 2
