Need algebra help with 2 x 2 system or linear equations that is inconsistent or consistent dependent
Algebra

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So, essentially what you want to do is turn this into one equation.
How you do this is by solving the first equation for a variable that you choose. Let's choose "x".
If I want to solve the first equation for x, I would need to isolate x on one side of the equation by itself. I would need to do the "inverse operations" to isolate the variable. This means that you need to do the opposite operation to get x by itself and get rid of whatever else is on the left side of the equation with x.
(Remember, we're still only on the first equation). These steps would be to add 3y to both sides of the equation, resulting in x = 3 + 3y.
x still isn't by itself because it is essentially being multiplied by 1. So the inverse operation would mean that you now have to divide both sides by 1. This will give you x = 3 + 3y. (Note that when you divide the right side of the equation by 1 both terms (the 3 and the 3y) change signs).
Now, you have solved the first equation for the variable x.
x = 3 + 3y.
Now that you know what the value of x is (in terms of y) you can substitute that value in for x in the second equation. So, the second equation is x + 3y = 3. Replace that x right there in the second equation with the value you got for x when you solved the first equation. So, when you solved the first equation you got that x = 3 + 3y. Just plug that 3 + 3y in for the x in your second equation.
The second equation is x + 3y = 3 and when you replace that x with the value you got in the first equation, you will now have (3 + 3y) + 3y = 3.
Combine like terms and solve. 3 + 3y + 3y = 3 becomes 3 = 3 because 3y and 3y added together equals zero. In this problem, you have now gotten rid of both variables and you have that 3=3. This will mean that the system has infinitely many solutions because you could substitute multiple values in for x and y in order to make the equations true.
On the second problem, if you solve the first equation for x you will get x = 5y 5
If you substitute that in for the x on the second equitation you will have (5y5)  5y = 5.
When simplified (because of distributing), this becomes 5y + 5  5y = 5
This can be simplified further to become 5= 5.
Now, 5 does not equal 5...this just isn't true. So, when you solve a system of equations and you simplify everything and you have eliminated the variables, but the answer doesn't end up being true like the first problem (where 3=3), this means that there is no solution to this system.
Please let me know if you need any clarification. I'm always happy to answer your questions.so in the 1st does y=3
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