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3 years ago

Solve the following equations:2x-5y-4 x+6y=15

Mathematics

1 answer:

Sauron [17]3 years ago

8 0

Answer:

x= 3

y= 2

Step-by-step explanation:

The given equations are :

2x-5y = -4 equation 1

x +6y = 15 equation 2

We will multiply the equation:2 by 2

2(x+6y=15)

2x+12y=30 equation 3

Lets call this equation:3

Now we will use the elimination method:

Subtract equation 3 from equation 1:

2x-5y = -4

2x+12y=30

_________

-17y = -34

y= -34/-17

y = 2

Now put the value of y in any equation:

We will use equation 1

2x-5y = -4

2x-5(2)= -4

2x-10= -4

Move the constant to the R.H.S

2x= -4+10

2x=6

Divide both the terms by 2

x= 3

Therefore the solution set is (x,y){(3,2)}....

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How do you solve this problem

Kryger [21]

Answer:

$sin(A) = \frac{21}{{29} }$

$cos(A)=\frac{20}{29 }$

$tan(A)=\frac{21}{20}$

$sin(C) = \frac{20}{{29} }$

$cos(C)=\frac{21}{29 }$

$tan(C)=\frac{20}{21}$

Step-by-step explanation:

So first we need to make sure we know the trig identities to solve this problem.

$sin(\alpha )=\frac{opposite}{hypotenuse}$
$cos(\alpha )=\frac{adjacent}{hypotenuse}$
$tan(\alpha )=\frac{opposite}{adjacent}$

Here we only have two legs of the triangle, so we will need to use the Pythagorean Theorem a² + b² = c² to solve for the missing leg, the hypotenuse in this case.

Solving for the hypotenuse, c, we get $c = \sqrt{a^{2} +b^{2} }$
Here a = 20 and b = 21, so plugging in these values to the equation we get: $c= \sqrt{(20)^{2}+(21)^{2} } =\sqrt{400+441} =\sqrt{841}=29$