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Sagot :
You can solve using substitution for system of equations.
So you have
y = 6x
and y = x^2 + 5
But it is also true that
y = y
since they are the same variable.
That means both y value equations are equal too:
y = 6x = y = x^2 + 5
So
6x = x^2 +5
Solve for x.
x^2 - 6x + 5 = 0
Factor to
(x-1)(x-5) = 0
x-1 = 0, x = 1
x-5 = 0, x = 5
So the two x-coordinate values will be (1,y value) and (5,y value)
Now to find y we use the original equations and we plug in the x values we found.
y = 6(1) = 6
y = (1)^2 + 5 = 6
So when x = 1, y = 6,
the coordinate is (1,6)
y = 6(5) = 30
y = (5)^2 + 5 = 30
When x = 5, y = 30
The coordinate is (5,30)
The answer is (1,6) and (5,30), which is A
So you have
y = 6x
and y = x^2 + 5
But it is also true that
y = y
since they are the same variable.
That means both y value equations are equal too:
y = 6x = y = x^2 + 5
So
6x = x^2 +5
Solve for x.
x^2 - 6x + 5 = 0
Factor to
(x-1)(x-5) = 0
x-1 = 0, x = 1
x-5 = 0, x = 5
So the two x-coordinate values will be (1,y value) and (5,y value)
Now to find y we use the original equations and we plug in the x values we found.
y = 6(1) = 6
y = (1)^2 + 5 = 6
So when x = 1, y = 6,
the coordinate is (1,6)
y = 6(5) = 30
y = (5)^2 + 5 = 30
When x = 5, y = 30
The coordinate is (5,30)
The answer is (1,6) and (5,30), which is A
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