Question

I have a very stupid question. I'm quite old and haven't done algebra since more than 15 years... now I stumbled upon this problem which I cannot solve. I feel ashamed. Can someone help me ?

Here it is:

y = x + 2
z = x +3
y = 2z + 1

Note: I don't want the answer, only a clear explanation of steps to use to resolve the problem. Thank you very much. :)

Answer 1

I'm 32, haha...

Answer 2

Old related to how long it's been since I last checked an algebra problem that is. ;)

Answer 3

You could use the elimination method.

Answer 4

I don't have a clue of whatever method whatsoever. I want a clear explanation with steps. :) No answer please.

Answer 5

y = 2z + 1
y = 2(x +3) + 1
y = 2((y-2) +3) + 1

y=2(y+1)+1
y=2y+3
y=-3/2

Answer 6

Oh my, I wasn't expecting a solution so soon, haha... :) I'll study those answers carefully. Thank you very much to all contributors, I love you, haha! :)

Answer 7

You have a system of three equations and you have three unknowns so theoretically this system can be solved. It is worth trying lol.
Take these two y=x+2 and y=2z+1 since y=y then we can say x+2=2z+1 obtaining a new equation x=2z-1.

Answer 8

Thank you very much everyone.

Answer 9

UnkleRhaukus used substitution
y = x + 2 (1)
z = x +3 (2)
y = 2z + 1 (3)
I would sub that expression for z in equation (2) into equation (3), which will give y in terms of x. between that and equation (1) you should be able to get a number value for x or y. From there on it is a system of two variables, so you can probably take it from there.
Good luck!

Answer 10

Now I want to come up with another equation that has only the variables x and z how about z=x+3 (given) from that we get x=z-3. Now we try to eliminate a variable.
x=z-3 and x=2z-1
x=2z-1 subtract
0=-z-4
z=-4 from there you substitute finding the others.

You manipulate the equations to eliminate a variable, then solve for an unknown and the follow through with substitution. By the way I am 73, and that it "senior"

Answer 11

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