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monitta
3 years ago
13

Use elimination method to solve 5x+y=-10 7x-3y=-36

Mathematics
1 answer:
k0ka [10]3 years ago
7 0
5x+y=-10\\ \\ 7x-3y=-36

Let's multiply the first equation by 3. (As you can see y's coefficient in the first one is 1 and in the second one is -3 , we will multiply the first equation by 3 so when we add the equations their sum will be 0)

3\cdot (5x+y)=3\cdot -10\\ \\ 15x+3y=-30

Now let's add this new equation and our second equation.

7x-3y=-36\\ \\ 15x+3y=-30\\ \\ (15x+3y)+(7x-3y)=(-30)+(-36)\\ \\ 15x+7x+3y-3y=-30-36\\ \\ 22x=-66\\ \\ x=\frac { -66 }{ 22 } \\ \\ x=-3

We found x=-3

Now let's plug x's value in one of the equations to find y's value.

x=-3\\ \\ 5x+y=-10\\ \\ 5\cdot -3+y=-10\\ \\ -15+y=-10\\ \\ y=-10+15\\ \\ y=5

So we found y=5

Solution ;

(-3, 5)

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The bearing of a lighthouse from a ship is N 37° E. The ship sails 2.5 miles further from the lighthouse. The new bearing is 25°
djverab [1.8K]

Answer:

The distance between the ship at N 25°E and the lighthouse would be 7.26 miles.

Step-by-step explanation:

The question is incomplete. The complete question should be

The bearing of a lighthouse from a ship is N 37° E. The ship sails 2.5 miles further towards the south. The new bearing is N 25°E. What is the distance between the lighthouse and the ship at the new location?

Given the initial bearing of a lighthouse from the ship is N 37° E. So, \angle ABN is 37°. We can see from the diagram that \angle ABC would be 180-37= 143°.

Also, the new bearing is N 25°E. So, \angle BCA would be 25°.

Now we can find \angle BAC. As the sum of the internal angle of a triangle is 180°.

\angle ABC+\angle BCA+\angle BAC=180\\143+25+\angle BAC=180\\\angle BAC=180-143-25\\\angle BAC=12

Also, it was given that ship sails 2.5 miles from N 37° E to N 25°E. We can see from the diagram that this distance would be our BC.

And let us assume the distance between the lighthouse and the ship at N 25°E is AC=x

We can apply the sine rule now.

\frac{x}{sin(143)}=\frac{2.5}{sin(12)}\\ \\x=\frac{2.5}{sin(12)}\times sin(143)\\\\x=\frac{2.5}{0.207}\times 0.601\\ \\x=7.26\ miles

So, the distance between the ship at N 25°E and the lighthouse is 7.26 miles.

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