To find out at what values of the parameter p the equation px – 3py + 6 = 0 is a pair of numbers (1.5; – 1.5), we substitute it into the equation x = 1.5 and y = – 1.5 and solve the resulting linear equation with one variable.
So, we substitute and get
1.5p – 3p * (- 1.5) + 6 = 0;
1.5p + 4.5p + 6 = 0;
We transfer the terms without a variable to the right side of the equation, while changing their sign to the opposite.
1.5p + 4.5p = – 6;
Here are similar ones on the left side of the equation:
6p = – 6;
Divide both sides of the equation by 6:
p = – 6: 6;
p = – 1.
Answer: at p = – 1.
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