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Alecsey [184]
3 years ago
7

Solve by substitution. 3x-y=5 -2x+y=0

Mathematics
2 answers:
Jobisdone [24]3 years ago
7 0
First, rewrite the equation -2x+y=0 into y-mx+b format. This is y=2x+0. We can then substitute this into our first equation.
3x-y=5
turns into
3x-2x=5
x=5
We can then plug in our value for x to find y.
3(5)-y=0
simplifies into
15-y=0
Simplify this to get that y=15
You can then plug these into the other equation ot make sure it's right.
-2(5)+15=5
-10+15=5
which is 5=5 (true)

So your answers are x=5 and y=15.

:)
Citrus2011 [14]3 years ago
3 0
In order to use substitution, you need to isolate a variable on one of the equations:

-2x+y=0
y=2x

Then you plug that equation into the other equation. Since we isolated y, we plug in 2x for y in the other equation:

3x-(2x)=5
x=5

Then plug the value you found into any equation to solve for the other variable:

3(5)-y=5
15-y=5
-y=-10
y=10

So your answers are x=5, y=10
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1. (a). A police car, approaching right-angled intersection from the north, is chasing a speedmg
tamaranim1 [39]

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<em>The SUV is running at 70 km/h</em>

Step-by-step explanation:

<u>Speed As Rate Of Change </u>

The speed can be understood as the rate of change of the distance in time. When the distance increases with time, the speed is positive and vice-versa. The instantaneous rate of change of the distance allows us to find the speed as a function of time.

This is the situation. A police car is 0.6 Km above the intersection and is approaching it at 60 km/h. Since the distance is decreasing, this speed is negative. On the other side, the SUV is 0.8 km east of intersection running from the police. The distance is increasing, so the speed should be positive. The distance traveled by the police car (y) and the distance traveled by the SUV (x) form a right triangle whose hypotenuse is the distance between them (d). We have:

d=\sqrt{x^2+y^2}

To find the instant speeds, we need to compute the derivative of d respect to the time (t). Since d,x, and y depend on time, we apply the chain rule as follows:

\displaystyle d\ '=\frac{x.x'+y.y'}{\sqrt{x^2+y^2}}

Where x' is the speed of the SUV and y' is the speed of the police car (y'=-60 km/h)

We'll compute  :

d=\sqrt{(0.8)^2+(0.6)^2}=\sqrt{0.64+0.36}

d=1\ km

We know d'=20 km/h, so we can solve for x' and find the speed of the SUV

\displaystyle \frac{x.x'+y.y'}{\sqrt{x^2+y^2}}=20

Thus we have

x.x'+y.y'=20

Solving for x'

\displaystyle x'=\frac{20-y.y'}{x}

Since y'=-60

\displaystyle x'=\frac{20+0.6(60)}{0.8}

x'=70\ km/h

The SUV is running at 70 km/h

7 0
3 years ago
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