SOLUTION: I have no idea how to do these, but its solving linear systems.
1) 35x+y=20
1.5x-0.1y=18
2) x+(1/3y)=-2
-8x-(2/3y)=4
Would you mind showing the work? as st
Algebra.Com
Question 623270: I have no idea how to do these, but its solving linear systems.
1) 35x+y=20
1.5x-0.1y=18
2) x+(1/3y)=-2
-8x-(2/3y)=4
Would you mind showing the work? as stated before, I have no idea how to do these, so anything to help me understand would be much appreciated. Thank you!
Found 3 solutions by lwsshak3, smmoore, MathTherapy:
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
I have no idea how to do these, but its solving linear systems.
1) 35x+y=20
1.5x-0.1y=18
multiply 2nd equation by 10 so the coefficient of y=that of the first equation
35x+y=20
15x-y=180
add
50x=200
x=4
plug in x=4 into first equation to solve for y
35*4+y=20
140+y=20
y=-120
check answers using 2nd equation
15*4-(-120)=60+120=180 (ok)
..
2) x+(1/3y)=-2
-8x-(2/3y)=4
multiply 1st equation by 2 so the coefficient of y=that of the 2nd equation
2x+(2/3y)=-4
-8x-(2/3y)=4
add
-6x=0
x=0
use 2nd equation to solve for y
0-2y/3=4
-2y=12
y=-6
check answers using 1st equation
0+2*-6/3=-4 (ok)
Answer by smmoore(10) (Show Source): You can put this solution on YOUR website!
okay i love doing these so hopfully this helps.
First of all when solving a system of linear equations what you are trying to do is to find where the two equations intersect or where their lines cross eachother. for the first set i think the easiest way to solve them is by substitution, which is where you get one letter by its self and then plug that new equation into the other one. then you take the value that you found and plug it back into one of the equations to find the other missing value.
35x+y=20
1.5x-0.1y=18
first i want to take the first equation and get y alone:
35x+y=20
-35x -35x
this gives you y= -35x+20
the next step is to put -35x+20 in for where ever you see y in the second equation.
so that means that: 1.5x-0.1(-35x+20)=18 then solve it
1.5x+3.5x-2=18 combine like terms and solve
5x-2=18
+2 +2 that gives you 5x=20 then divide to get x alone ; 5x/5=x and 20/5=4 so x=4
then the next step is to put 4 in place of x in one of the original equations. i think the first one is easier, but you can choose either one.
35x+y=20 so putting 4 in for x i get: 35(4)+y=20 then solve
140+y=20
-140 -140 that gives me:
y= -120
after solving for x and y i know that the solution is (4,-120) which is the coordinates of the intersecting point, or (x,y)
but sometimes you can just put what x equals and what y is seperately: x=4 and y=-120
for #2 equations i will do the exact same thong but with the different numbers.
this time i will get x alone first.
x+(1/3y)=-2
-8x-(2/3y)=4
x+(1/3y)=-2
-(1/3)y -(1/3)y which equals: x= -(1/3)y-2
Put that un for x in the second equation
-8(-(1/3)y-2)-(2/3)y=4 solve and combine like terms
(8/3y)+16-(2/3y)=4 (8/3y)-(3/3y)=(5/3y)
(5/3y)+16=4
-16 -16
(5/3y)= -12 5/3y /5/3=y -12/(5/3)= -7.2
y= -7.2
we have the y value, now put -7.2 in for y in the first equation
x+(1/3)*(-7.2)= -2
x-2.4= -2
+2.4 +2.4
x=0.4
x= 0.4 and y= -7.2
but the coordinate answer is (0.4,-7.2)
Answer by MathTherapy(10552) (Show Source): You can put this solution on YOUR website!
I have no idea how to do these, but its solving linear systems.
1) 35x+y=20
1.5x-0.1y=18
2) x+(1/3y)=-2
-8x-(2/3y)=4
Would you mind showing the work? as stated before, I have no idea how to do these, so anything to help me understand would be much appreciated. Thank you!
No. 1
It's always best to rid the equation of decimals
35x + y = 20 ---- eq (i)
1.5x - 0.1y = 18 ---- eq (ii)
15x - y = 180 -- Multiplying eq (ii) by 10 to rid it of its decimals --- eq (ii)
35x + y = 20 --- eq (i)
50x = 200 ----- Adding eqs (ii) & (i)
x = , or
35(4) + y = 20 ------ Substituting 4 for x in eq (i)
140 + y = 20
y =
No. 2
It can be very ugly working with fractions, so the best thing to do is get rid of them
------ eq (i)
----- eq (ii)
6x + 2y = - 12 ----- Multiplying eq (i) by 6 ----- eq (iii)
- 24x - 2y = 12 ----- Multiplying eq (ii) by 3 ----- eq (iv)
- 18x = 0 ----- Adding eqs (iii) and (iv)
x =
0 + 2y = - 12 ----- Substituting 0 for x in eq (iii)
2y = - 12
y = , or
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