David M. answered • 04/08/16
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Math n Physics - Top Credentials, Real-World Exp, Cert'd, Open Sched
H Minah.
There are multiple ways to solve systems of equations. Before jumping in with one method, it probably will be benificial to look at the systems for clues as to which method might be easiest. I'll look at the first system. All equations are in order x, y, z. Each equation has all variables. Substitution is probably not the best approach. Would be messy to go that route. The top two equations has opposite coefficieints for z (-1 amd 1). So we can use elimination there:
x + 2y - z = 5
+(2x - 4y +z) = 0
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3x -2y +0 = 5
We can also multiply the 2nd eqn by -2 so the coeff of z will become -2 (opposite the 2z) in the third equn. So we can use this strategy to use the 2nd and third eqn to eliminate z again and end up with another eqn with just x & y.
-2(2x -4y + z = 0 ) -> -4x+8y-2z = 0
3x +2y+2z=3 (now let's add the third eqn like we did above)
_____________
-x +10y +0 = 3
So we can now use these two equations with x & y to solve for x and y.
3x-2y=5
-x+10y=3 (lets continue using elimination. I like to keep things simple and recognize by multiplying the bottom eqn by 3, the coefficients for x will be opposites (3 and -3) and thus, they will cancel if we add the two equations (allowing us to solve for y.
3x -2y = 5
-3x +30y= 9
__________
28y=14
y= 1/2 or .5
Now we can plug this back into one of the x,y equations we just used. Top one is easy enough
3x -2(1/2)=5
3x -1=5
3x=6
x=2
Now we need to go back to one of the original three equations and sub in our solutions for x and y which will only leave z. The top equation looks pretty simple, so let's try that
x+ 2y-z =5
2 + 2(1/2) - z =5
2+1 -z =5
3-z=5
3-5=z
-2=z
Ans (2, 1/2, -2)
Check: lets try them in Eqn 3: 3(2) +2(1/2) +2(-2)=3
6 + 1 -4 = 3
3= 3 Check of this equation works.
