Question

USING THE ELIMINATION METHOD,SOLVE THE SYSTEM OF EQUATIONS:

A+9B=11
A+2B= -3

Answer 1

take the second equation from the first

Answer 2

solve for B, then back substitute to find A

Answer 3

\[A+9B=11\qquad\qquad{\small(i)}\]\[A+2B= -3\qquad\qquad{\small(ii)}\]

\[(A+9B)-(A+2B)=11-(-3)\qquad\qquad{\small(i)-(ii)}\]

Answer 4

SO THE ANSWER IS 2,11

Answer 5

to check your answer plug into one of the original equations
to see if B=2, A=11 is right ;

\[A+9B=11\qquad\Longrightarrow 11+9\times2=11?\]

Answer 6

B=2 &A =11 RIGHT

Answer 7

that dosent work

Answer 8

RIGHT

Answer 9

\[(A+9B)-(A+2B)=11-(-3)\qquad\qquad{\small(i)-(ii)}\]\[A-A+9B-2B=11+3\]\[7B=14\]\[B=2\]

Answer 10

\[A+9\times(2)=11\]\[A=\dots?\]

Answer 11

2

Answer 12

B is two , you have got this right,

but what is A?

Answer 13

A=2,B=2, dosent work because

\[2+9\times2=20\neq11\]

Answer 14

SO IM LOST..

Answer 15

you have found B right?

Answer 16

RIGHT

Answer 17

but i thought the way i was doinit was correct...guess not :(

Answer 18

there are heaps of ways to solve this sort of problem,
if you'd rather a different method thats fine but you'll have to let me know what it is,
if you want some help

we have \(B=2\) so looking back at the original equations

\[A+9B=11\qquad \Rightarrow A+9(2)=11\]
\[A+2B= -3\qquad \Rightarrow A+2(2)= -3\]

all we need is A

Answer 19

Using the first equation we could find A like this:\[A+9\times2=11\]\[A+18=11\]\[A=11-18\]
Or using the second equation
\[A+2\times2=−3\]\[A+4=-3\]\[A=-3-4\]

Answer 20

which ever method you use you'll find that A is .....

Answer 21

-7

Answer 22

if you are right both these should be true
(ie LHS=RHS)

(-7)+9(2)=11

(-7)+2(2)= -3

Answer 23

what are the left hand sides (LHS's)

Answer 24

-7+9x2=-7+18=...

-7+2x2=-7+4=...
/?

Answer 25

what do you get?

dose A=-7, B=2 work for both of the original equations?

Answer 26

if so, you have solved the system