Solve using the elimination method. Show your work. If the system has no solution or an infinite number of solutions, state this. 2x + 10y = 58 -14x + 8y = 140
hey @RenaeMonroe . do you have any idea about elimination method?
well i think i just divide by 2 but not sure
Ok, I'll explain you
We have two equations \[2x + 10y = 58 \]\[-14x + 8y = 140 \] Here we'd eliminate one of the variables either x or y by adding or subtracting the two equations let's see which one is easy to eliminate. if I choose y, coefficients of y in the two equations 10 and 8 so I'll multiply the first equation by 4 and second equation by 5 \[4\times(2x+10y)=58 \times 4=>8x+40=232\] \[5\times(-14x + 8y )= 140\times 5 =>-70x+40y=700\] so the coefficients of y in both will become 40 do you get this part?
yes
Now we'll subtract the two equations, y will get eliminated and then we could find x. Then we'll plugin the value of x in any of the original equations to find y. Do you understand this?
8x+40y=232 - -70x+40y=700 ------------------ -78+y=468
|dw:1347344074078:dw|
do you get this?
yes
2(-6)+10y=58
-12+10y=58 10y=58-12 10y=46 divide 46/10=4.6 so y = 4.6 is this right
a small mistake \[-12+10y=58\] Add 12 to both the sides \[-12+12+10y=58+12\] \[10y=70\]
ok ic so y = 7
@RenaeMonroe did you understand the way?
yes i just forgot about that step
ok, now would you try to solve this problem eliminating x ?
14(2x+10y)58*14=>28x+140y=812 2(-14x+8y)=140*2=>-28x+16y=280 28x+140y=812 +(-28x)+16y=280 =156y=1092 y=7
very good but there's a short way here multiply first by 7 and then just add it to the second \[14x+70y=406\] \[-14x+8y=140\] \[78y=546\]
oh k thanks
welcome :D
thanks for the medal to
:D
i have a few more questions to get answers for.
close this and post them one by one :)
ohk thanks
Join our real-time social learning platform and learn together with your friends!