Alg2 help (picture) Elimination
Nvm I get this one
2x + y = 9 x - 2z = -3 2y + 3z = 15 Is this the problem you need to solve by elimination?
#5
Nvm I actually need help in both...
2x + y = 9 ----- (1) x - 2z = -3 ----- (2) 2y + 3z = 15 -----(3) Multiply second equation by 2 and subtract it from the first equation: 2x + y = 9 2x - 4z = -6 Subtract y + 4z = 15 Multiply the above equation by 2 and subtract it from equation 3 2y + 8z = 30 2y + 3z = 15 Subtract 5z = 15 z = 3 2y + 3z = 15. Substitute z = 3 2y + 9 = 15 2y = 15 - 9 = 6 y = 3 Equation 1) is 2x + y = 9. Substitute y = 3 2x + 3 = 9 2x = 9 - 3 = 6 x = 3 So x = 3, y = 3, z = 3
You can use the same method to solve the other equation: 2x - y + z = -4 ------- (1) 3x + y - 2z = 0 ------- (2) 3x - y = -4 ------- (3) The last equation has only x and y and no z. So try eliminating z using equations 1 and 2 first. Let me know if you get stuck.
alright Ill try it now!
I got 7x - y = - 8
I multiplied 1st equation by 2
then eliminated with the second equation
Correct. So you have 7x - y = -8 and the third equation is: 3x - y = -4 Eliminate y from these two equations.
Ohh.. I just used that equation and made y = 7x + 8 then I substituted in 3x - y = -4
Is that correct too?
You can do that too. But simply subtracting the two equations will eliminate y 7x - y = -8 3x - y = -4 subtract to get rid of y.
So far I got x is 3 and y is 13
z is 3
Not correct.
all of them?
In these types of problems if you get one of the values wrong then all other values will also be wrong because they depend on the the other values.
Okay so how would you do it?
Well, let us retrace your steps. Earlier you had y = 7x + 8 and you said you substituted y in 3x - y = -4 Why don't you do that here and I will see what you get.
x is 3
No. Show me the step-by-step before you solved for x. Start with y = 7x + 8 and substitute y in 3x - y = -4 Show me the step-by-step
\[2x - y + z = -4 \]
I multiplied it by 2
\[7x - y = -8\]
This is what I got, then I subtracted 7x from both sides and I got:
\[y = 7x +8\]
OK. Next step?
Alright so then I plugged in the Y for: \[3x - 7x + 8 = -4\]
3x - 7x is = -4
then minus 8 from both sides and -4x = -12..so now divide by -4 and x is 3
The mistake is in plugging in the y.
positive 7?
You had y = 7x + 8 Plug y in 3x - y = -4 3x - (7x + 8) = -4 3x -7x - 8 = -4 (you had +8 and that is the error)
why is 8 negative?
y = 7x + 8 When you put the above y in 3x - y you have to remember to put a parenthesis around y so it should be 3x - ( substitute the y value here ) 3x - (7x + 8) Now remove the parentheses. 3x - 7x - 8
3x - y means you are subtracting y from 3x. But if y is made up of two parts: 7x + 8 then each part of y has to be subtracted from 3x so it becomes 3x - 7x - 8
ohh okay!
then how do I combine 3x - (7x+8) = -4 ?
what is -(a + b) if you remove the parenthesis?
-(a + b) = -a - b
x is 1?
so -(7x + 8) = -7x - 8
Not quite. X is not 1. Watch out for the signs
7x - 8?
You have 3x - (7x + 8) = -4 First get rid of the parenthesis. When ridding the parenthesis, the minus sign outside should be applied to each term inside the parenthesis. So 3x - (7x + 8) = -4 becomes 3x - 7x - 8 = -4
i got 1 for x again
what is 3x - 7x ?
-4x
correct. so -4x - 8 = -4 solve for x
plus 8 both sides
corrcet.
\[-4x = -4\]
then divide by -4
which is 1
-4x - 8 = -4 8 8 (add 8 to both sides of the equation) -4x = 4 (not minus 4) So x = ?
ohh!! -1 !!
Good!
What is y & z?
x = -1 y = 1 z = 1
check for z again.
z = -1
Yes! x = -1, y = 1, z = -1. Once you solve these type of problems always substitute the values for x, y, z into the left hand side of the original 3 equations to see if you get the right hand side. That way you will know your answers are correct.
FINALLY!!!!
thank you for helping me!
No problem. Pay attention to the signs. Use parenthesis when substituting when there are many terms. Good Luck!
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