SOLVE X/2+2Y/3=-1 AND X-Y/3=3USING SUBSTITUTION AND ELIMINATION METHOD
To solve the system of equations using the substitution method, we'll solve one equation for one variable and substitute it into the other equation. Let's start with the first equation:
1) X/2 + (2Y/3) = -1
To eliminate the fractions, we can multiply every term in the equation by the least common denominator (LCD), which is 6:
6 * (X/2) + 6 * (2Y/3) = 6 * (-1)
2X + 4Y = -6
Now, let's solve the second equation:
2) X - (Y/3) = 3
To eliminate the fraction, we'll multiply every term in the equation by the LCD, which is 3:
3 * X - 3 * (Y/3) = 3 * 3
3X - Y = 9
Now we have a new system of equations:
2X + 4Y = -6
3X - Y = 9
Now, we can solve the first equation for X:
2X = -6 - 4Y
Divide by 2 on both sides:
X = (-6 - 4Y)/2
Simplifying further:
X = -3 - 2Y
Now we substitute this expression for X into the second equation:
3(-3 - 2Y) - Y = 9
Distribute 3 to both terms inside the parentheses:
-9 - 6Y - Y = 9
Combine like terms:
-7Y = 18
Divide by -7:
Y = -18/7
Now that we have the value of Y, we can substitute it back into equation (1) to find the value of X:
X = -3 - 2(-18/7)
X = -3 + 36/7
X = (49 - 36)/7
X = 13/7
Therefore, the solution to the system of equations is X = 13/7 and Y = -18/7.
x/2 + 2y/3 = -1 ---> 3x + 4y = -6
x - y/3= 3 --------> 3x - y = 9
elimination:
subtract them:
5y = -15
y = -3
sub into the 2nd
3x - (-3) = 9
3x = 6
x = 2
x = 2, y = -3
by substitution:
from the 2nd: y = 3x-9
into the 1st:
3x + 4(3x-9) = -6
3x + 12x - 36=-6
15x = 30
x = 2
back into y = 3x-9 = 3(2) - 9 = -3
x = 2, y = -3