Solve by substitution:
x-y=3
x+3y=6
This is what I get:
X+3y=6
Y=x-3
X(x-3)=6
2x= 9
2 2
X= 9/2
9/2+3y=6
9/2-9/2+3y= 6-9/2
3y/3= 3/2/3
Y= ½
man, ya gotta check your answer. It does not fit the first equation.
y = x-3, so
x + 3(x-3) = 6
4x = 15
x = 15/4
y = 3/4
thank you
To solve this system of equations by substitution, follow these steps:
Step 1: Solve one of the equations for one of the variables. Let's choose the first equation and solve it for x:
x - y = 3 (equation 1)
x = y + 3 (add y to both sides)
Step 2: Substitute the expression found for x in step 1 into the other equation. Let's substitute x = y+3 in the second equation:
x + 3y = 6 (equation 2)
(y+3) + 3y = 6 (substitute x)
4y + 3 = 6 (combine like terms)
Step 3: Solve for y:
4y = 6 - 3 (subtract 3 from both sides)
4y = 3
y = 3/4 (divide both sides by 4)
Step 4: Substitute the value of y back into equation 1 to solve for x:
x = y + 3
x = (3/4) + 3
x = 3/4 + 12/4 (convert 3 to 12/4 for like denominators)
x = 15/4
So, the solution to the system of equations is x = 15/4 and y = 3/4.