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how to solve each system by elimination 16x-10y=10 and -8x-6y=6

i dont know how to do this
please explain
thanks
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3 Answers

Hi again Angelica;
16x-10y=10 and -8x-6y=6
The coefficients of x are 16 and -8. To apply elimination, these should be the same.  Positives and negatives are non-issues.
Let's take the first equation...
16x-10y=10
Let's multiply both sides by 1/2...
(1/2)(16x-10y)=(10)(1/2)
8x-5y=5
Let's add this to the second equation...
-8x-6y=6
8x-5y=5
0-11y=11
The x is eliminated...
-11y=11
Let's divide both sides by -11...
(-11y)/11=11/-11
y=-1
Let's plug this into either original equation to establish the value of x.  I randomly select the second...
-8x-6y=6
-8x-[(6)(-1)]=6
-8x-(-6)=6
Subtracting a negative number is identical to adding a positive number...
-8x+6=6
Let's subtract 6 from both sides...
-8x+6-6=6-6
-8x=0
x=0
Let's plug both values into the first equation to verify results...
16x-10y=10
[(16)(0)]-[(10)(-1)]=10
0-(-10)=10
10=10
 
Solving by elimination with two linear equations can be most easily described as adding the two together. The goal is to eliminate either the x or y variable, then solve for the remaining variable. Since neither coefficient are additive inverses ( 16x + -8x =/= 0), the easiest remedy is to multiple both sides of the second equation by 2.
 
2 (-8x - 6y) = 2(6) results in -16x - 12y = 12
 
Now, add this to the first equation 16x - 10y = 10:
 
     16x - 10y = 10
+  -16x - 12y = 12
=    0x  - 22y = 22
 
Solve for y:
-22y = 22   divide both sides by -22
    y  = -1
 
Now that you know y = -1, substitute it back into the original equation:
16x - 10 (-1) = 10
16x + 10 = 10
16x = 0
  x  = 0
 
To check, substitute x = 0 into the second original equation:
-8 (0) - 6y = 6
-6y = 6
y = -1
 
I hope this is what you were looking for!
see the answers to the prior 4 or 5 questions posed by you on solving by elimination