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Solve Simultaneously 2x+y =3, 4х2- у2+2x+3y=16​

User Asonge
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1 Answer

4 votes

Answer:

x = 1/13 and y = 37/13.

Explanation:

To solve the system of equations:

2x + y = 3 ...(1)

4x^2 - y^2 + 2x + 3y = 16 ...(2)

We can use the method of substitution or elimination.

Let's start by using the substitution method. Rearrange equation (1) to solve for y:

y = 3 - 2x

Now substitute this expression for y in equation (2):

4x^2 - (3 - 2x)^2 + 2x + 3(3 - 2x) = 16

Expand and simplify the equation:

4x^2 - (9 - 12x + 4x^2) + 2x + 9 - 6x = 16

Combine like terms:

4x^2 - 9 + 12x - 4x^2 + 2x + 9 - 6x = 16

Simplify further:

4x + x - 6x - 12x = 16 - 9 - 9 + 6

Combine like terms:

-13x = -1

Divide both sides of the equation by -13:

x = 1/13

Now substitute this value of x back into equation (1) to solve for y:

2(1/13) + y = 3

Multiply through by 13 to eliminate the fraction:

2 + 13y = 39

Subtract 2 from both sides:

13y = 37

Divide both sides by 13:

y = 37/13

Therefore, the solution to the system of equations is:

x = 1/13

y = 37/13

So, the values of x and y that satisfy both equations are x = 1/13 and y = 37/13.

User DSchultz
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