Question

X²+4x+4+y²-2y+1-x²-y²=5 3y+x=28​

Answer 1

1275/214=x. 89/214=y

Answer 2

It seems like you have a system of two equations involving variables x and y:

Equation 1: $x^2 + 4x + 4 + y^2 - 2y + 1 - x^2 - y^2 = 5$

Equation 2: $3y + x = 28$

Let's simplify and solve this system of equations step by step:

Equation 1:
Simplify the terms and cancel out the $x^2$ and $y^2$ terms:

$4x - 2y + 5 = 5$

This simplifies to:

$4x - 2y = 0$

Equation 2: This equation is already in a simplified form:

$3y + x = 28$

Now we have a system of equations:

$4x - 2y = 0$ $3y + x = 28$

There are different methods to solve this system, but let's use the substitution method:

Step 1: Solve Equation 2 for $x$:

$x = 28 - 3y$

Step 2: Substitute the expression for $x$ from Step 1 into Equation 1:

$4x - 2y = 0$ $4(28 - 3y) - 2y = 0$

Simplify:

$112 - 12y - 2y = 0$ $112 - 14y = 0$ $14y = 112$ $y = 8$

Step 3: Substitute the value of $y$ back into the expression for $x$ from Step 1:

$x = 28 - 3y$ $x = 28 - 3(8)$ $x = 28 - 24$ $x = 4$

So, the solution to the system of equations is $x = 4$ and $y = 8$.