Solve by using the substitution method
1.x=-3y+4 and 3x+9y=12
3(-3y+4)+9y=12
-9y+12+9y12
0=0
x=-3(0)+4
x=4
Would there be infinite solutions or would it be (0,4)?
2. x=3y-1 and 2x-6y=-2
2x-6y=-2
2(3y-1)-6y=-2
6y-2-6y=-2
0=0
x=3y-1
x=3(0)-1
x=-1
Would there be infinite solutions or would it be (-1,0)?
for your first one, the variable dropped out, but you ended up with a true statement, 0=0
If that happens you will have an infinite number of solutions. In effect , the two equations are really the same equation.
look at the 2nd
3x + 9y = 12 , divide by 3
x + 3y = 4
or
x = -3y + 4 , which was your first equation.
If your variable drops out, but you end up with a false statement, such as 4 = 9
then there is no solution.
In that case, what you would be given would be two lines that are parallel, thus they never meet, thus no solution
The same result was true in your second question
notice by dividing each term of the 2nd by 2, then re-arranging you get your first equation
I also noticed that in both cases you substituted x=0
but nowhere did it say that x = 0 , it said 0 = 0
no mention of x
So if you have a true statement, ---> infinite number of solutions
If you have a false statement --> no solution at all
Thank you so much.
welcome
In the first problem, you correctly substituted the value of x from the first equation into the second equation to get:
3(-3y+4) + 9y = 12.
However, when you simplified the equation, you made a calculation mistake. The correct simplification should be:
-9y + 12 + 9y = 12.
Here, the -9y and 9y terms cancel each other out, leaving you with:
12 = 12.
This equation is true for all values of y. Therefore, the system of equations has infinitely many solutions.
To find the value of x when y is 0, you correctly substituted y=0 into the equation x = -3y + 4, which gives x=4. So, one solution is (0, 4).
In the second problem, you also correctly substituted the value of x from the first equation into the second equation:
2(3y-1) - 6y = -2.
Simplifying this equation, you made a calculation mistake:
6y - 2 - 6y = -2.
Here, the 6y and -6y terms cancel each other out, leaving you with:
-2 = -2.
This equation is true for all values of y. Therefore, the system of equations has infinitely many solutions.
To find the value of x when y is 0, you correctly substituted y=0 into the equation x = 3y - 1, which gives x=-1. So, one solution is (-1, 0).
In both cases, the system of equations has infinitely many solutions, and the specific solution you provided is correct.