You want to isolate one of the variables (x or y) so you can plug it into the other equation. The easiest one is isolating the 2nd equation.
3x² - 16x + 13 - y = 0 Add y on both sides
3x² - 16x + 13 = y
You can use this and plug it into the first equation
y - 12x + 15 = 3x²
(3x² - 16x + 13) - 12x + 15 = 3x²
3x² - 16x + 13 - 12x + 15 = 3x² Combine like terms
3x² - 28x + 28 = 3x² Subtract 3x² on both sides
-28x + 28 = 0 Add 28x on both sides
28 = 28x Divide 28 on both sides
1 = x
Now that you know x, you can plug it into either of the equation to find y
3(1)² - 16(1) + 13 - y = 0
3 - 16 + 13 - y = 0
-y = 0 Divide -1 on both sides
y = 0
x = 1, y = 0
Answer:
b = 60
Step-by-step explanation:
Using Pythagoras' identity in the right triangle
The square on the hypotenuse is equal to the sum of the squares on the other 2 sides, that is
b² + 63² = 87²
b² + 3969 = 7569 ( subtract 3969 from both sides )
b² = 3600 ( take the square root of both sides )
b = 
= 60
Answer: Its C
Step-by-step explanation:
Answer:
b = 3a
Step-by-step explanation:
a = < - 1, 3 >
b = < - 3, 9 >
Thus
b = 3a
