Solve system by elimination y=x^2 y=x+2 SHOW YOUR WORK

1 answer:

3 0

We have that
y=x²----> equation 1<span>
y=x+2-----> equation 2

multiply equation 1 by -1
-y=-x</span>²

add equation 1 and equation 2
-y=-x²
y=x+2
------------
0=-x²+x+2-------------> -x²+x+2=0-----> x²-x-2=0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(x²-x)=2
<span>Complete the square. Remember to balance the equation by adding the same constants to each side
</span>(x²-x+0.5²)=2+0.5²

Rewrite as perfect squares

(x-0.5)²=2+0.5²
(x-0.5)²=2.25-----> (x-0.5)=(+/-)√2.25-----> (x-0.5)=(+/-)1.5
x1=1.5+0.5-----> x1=2
x2=-1.5+0.5---- > x2=-1

for x=2
y=x²----> y=2²----> y=4
the point is (2,4)

for x=-1
y=x²----> y=(-1)²---> y=1
the point is (-1,1)

the answer is
the solution of the system are the points
(2,4) and (-1,1)

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When truckers are on long-haul drives, their driving logs must reflect their average speed. Average speed is the total distance

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a) $v=\frac{d_{tot}}{t_{tot}}=\frac{(3 h)(60 mph)+20 mi}{3 h +t_2}$

The average speed is equal to the ratio between the total distance ( $d_{tot}$ and the total time taken ( $t_{tot}$ ):

$v=\frac{d_{tot}}{t_{tot}}$

the distance travelled by the trucker in the first 3 hour can be written as the time multiplied by the velocity:

$d_1 = (3 h)(60 mph)=180 mi$

So the total distance is

$d_{tot}=d_1 +d_2 = 180 mi+20 mi=200 mi$

The total time is equal to the first 3 hours + the time taken to cover the following 20 miles in the city:

$t_{tot}=3 h +t_2$

So, the equation can be rewritten as:

$v=\frac{d_{tot}}{t_{tot}}=\frac{(3 h)(60 mph)+20 mi}{3 h +t_2}$

b) 0.50 h (half a hour)

Since we know the value of the average speed, $v=57.14 mph$ , we can substitute it into the previous equation to find the value of $t_2$ , the time the trucker drove in the city: