Very basic question, but still cant find the answer!?

Question

Ok, the answers I keep getting don't match what is in the back of the book, even though I've gotten all the previous questions in the book right. So obviously, one of us is wrong (probably me), so I want you guys to tell me what the answers are. Here is the question:

"Solve the following simultaneous equations using the substitution method:"

3x + 2y = 33
y = 41 - 5x

So basically, using the substitution method, I need to find the values of the variables. Can anyone tell me the answers?

Answer 1

3x + 2y = 33
y = 41 - 5x

Substitute the second into the first:

3x + 2(41 - 5x) = 33
=> 3x + 82 - 10x = 33
=> -7x = -49
=> x = 7.
y = 41 - 5x = 41 - 35 = 6.
So x = 7 and y = 6.

Answer 2

3x + 2y = 33
y = 41 - 5x

Substitute the second into the first:

3x + 2(41 - 5x) = 33
=> 3x + 82 - 10x = 33
=> -7x = -49
=> x = 7.
y = 41 - 5x = 41 - 35 = 6.
So x = 7 and y = 6.

Answer 3

3x + 2y = 33---(1)
y = 41 - 5x-----(2)

substitute (2) into (1)

3x+2(41-5x)=33
therefore
3x+82-10x=33
3x-10x=33-82
-7x= -49
-x =-49/7
-x=-7
therefore x=7
substitute x=7 back into ---(2)
y = 41 - 5x
y = 41 - (5*7 )
y = 41 - 35
therefore y = 6

p.s
if u sub x=7 and y=6 back into 3x + 2y = 33
u should get 33
3x + 2y = 33
=3(7) + 2(6)
=21 + 12
therefore= 33
this prove answer is correct

Answer 4

you don't need the back of the book to check if your answer is right.
replace the values obtained in your solution and you will get an identity if it's Right.
in this case:
[x = 7 ∧ y = 6] are the solutions then replace this values as well:
first equation:
3(7) + 2(6) = 33
21+12=33
33=33 OK

second equation:
6 = 41 - 5(7)
6=41-35
6=6 OK

so the solution [x = 7 ∧ y = 6] is good.

Answer 5

3x+2y=33 ; y=41-5x substitute value of y then, 3x + 2 [41-5x]=33 now solve , 3x + 82 -10x=33 then, 82-33= 10x -3x thus, 49=7x then, x= 7 and y = 41-5*7 ans is y = 6

Answer 6

you replace the value of y in the first equation by the second expression
So : 3x+2*(41-5x)= 33
-7x+82 = 33
-7x= 33-82 =-49

x=7 and y =41-5*7 = 41-35 =6

you replace x by 7 and y by 6 and check that it is correct

Answer 7

a.) 3x + 2y = 33
y= 41 - 5x

substitute:
b.) 3x + 2(41 - 5x) = 33

solve:
c.) 3x + 82 - 10x = 33

move to one side:
subtract 82 from 33 = -49
d.) 3x - 10x = -49

subtract 10 from 3 = -7
e.) -7x = -49

divide -49 by -7 = 7
f.) x = 7

substitute (part 2):
g.) 3(7) + 2y = 33

solve:
multiply 3 by 7 = 21
h.) 21 + 2y = 33

move to one side:
subtracts 21 from 33 = 12
i.) 2y = 12

divide 12 by 2 = 6
j.) y = 6

check:

3(7) + 2(6) = 33
21 + 12 = 33

correct!



I probably got that one wrong, can I copy from someone else?

Answer 8

3x + 2(41 - 5x)= 33
3x + 82 - 10x = 33
-7x + 82 = 33
-7x = -49
x = 7
there you go!!!
y = 41 - 5(7)
y = 41 - 35
y = 6

Answer 9

i got x=7 and y=6 here's how i got it...
3x+2(41-5x)=33
3x+82-10x=33
-7x=-49
-7/-49=7

then...y=41-5(7) ---------> y=41-35 -----> y=6

Answer 10

You can solve it eaisly;

Put y = ... in eq 1; you get

3x + (41-5x) = 33
thus

7x = 81 - 33
7x = 52
thus

x = 52 / 7;

Substitute this is 2 you get

y = 41 - 5 * 52/7
thus

y = 0.28571428571428571428571428571429