To solve this system of equations, we can use the substitution method.
Starting with the first equation:
X - 2y = 5X = 5 + 2y
Now we can substitute this value of X into the second equation:
(5 + 2y)^2 + 2y = 5125 + 20y + 4y^2 + 2y = 514y^2 + 22y - 26 = 02y^2 + 11y - 13 = 0
Now we can factor this quadratic equation:
(2y - 1)(y + 13) = 0
Setting each factor to zero gives us two possible solutions for y:
2y - 1 = 02y = 1y = 1/2
And
y + 13 = 0y = -13
Now that we have the values of y, we can substitute them back into the equation X = 5 + 2y to find the values of X:
For y = 1/2:X = 5 + 2(1/2)X = 5 + 1X = 6
For y = -13:X = 5 + 2(-13)X = 5 - 26X = -21
Therefore, the solutions to the system of equations are X = 6, y = 1/2 and X = -21, y = -13.
To solve this system of equations, we can use the substitution method.
Starting with the first equation:
X - 2y = 5
X = 5 + 2y
Now we can substitute this value of X into the second equation:
(5 + 2y)^2 + 2y = 51
25 + 20y + 4y^2 + 2y = 51
4y^2 + 22y - 26 = 0
2y^2 + 11y - 13 = 0
Now we can factor this quadratic equation:
(2y - 1)(y + 13) = 0
Setting each factor to zero gives us two possible solutions for y:
2y - 1 = 0
2y = 1
y = 1/2
And
y + 13 = 0
y = -13
Now that we have the values of y, we can substitute them back into the equation X = 5 + 2y to find the values of X:
For y = 1/2:
X = 5 + 2(1/2)
X = 5 + 1
X = 6
For y = -13:
X = 5 + 2(-13)
X = 5 - 26
X = -21
Therefore, the solutions to the system of equations are X = 6, y = 1/2 and X = -21, y = -13.