Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
5. Systems With Three Variables
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Exercise 38 Page 172

When using the Substitution Method to solve a system of equations, it is necessary to isolate a variable.

(72/10,216/10,144/10)

Practice makes perfect
The given system consists of equations of planes. We will use the Substitution Method to solve this system because in the second and the third equations there are isolated variables. The first step will be to substitute the values of l and h into the first equation and solve for w.
2l+2w+h=72 l=3w h=2w
2( 3w)+2w+ 2w=72 l=3w h=2w
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(I):Solve for w
6w+2w+2w=72 l=3w h=2w
10w=72 l=3w h=2w
w= 7210 l=3w h=2w
The value of w is 7210. Substituting 7210 for w into the second and the third equations, we can find the values of l and h.
w= 7210 l=3w h=2w

(II), (III): w= 7210

w= 7210 l=3( 7210) h=2( 7210)

(II), (III): a*b/c= a* b/c

w= 7210 l= 3(72)10 h= 2(72)10

(II), (III): Multiply

w= 7210 l= 21610 h= 14410
The solution to the system is the point ( 7210, 21610, 14410). This is the singular point at which all three planes intersect.