Exercise 23 Page 172

The given system consists of equations of planes. When using the Substitution Method to solve a system of equations, it is necessary to isolate a variable. In the third equation, r is already isolated, so we can substitute its equivalent expression into the remaining equations. In the final step of the simplification of these substitutions, our goal is to have yet another variable isolated. This time, the t-variable was isolated in the second equation. We can now substitute its equivalent expression into the first equation.

11s+5-3t=3 t=4s+1 r=3s+1

Substitute

(I): t= 4s+1

11s+5-3( 4s+1)=3 t=4s+1 r=3s+1

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(I): Solve by substitution

Distr

(I): Distribute (-3)

11s+5-12s-3=3 t=4s+1 r=3s+1

SubTerms

(I): Subtract terms

- s+2=3 t=4s+1 r=3s+1

SubEqn

(I): LHS-2=RHS-2

- s=1 t=4s+1 r=3s+1

MultEqn

(I): LHS * (-1)=RHS* (-1)

s=-1 t=4s+1 r=3s+1

The value of s is -1. Substituting -1 for s into the second and the third equations, we can find the values of t and r.

s=-1 t=4s+1 r=3s+1

(II), (III): s= -1

s=-1 t=4( -1)+1 r=3( -1)+1

(II), (III): a(- b)=- a * b

s=-1 t=-4+1 r=-3+1

(II), (III): Add terms

s=-1 t=-3 r=-2

The solution to the system is the point ( -2, -1, -3). This is the singular point at which all three planes intersect. Let's check our solution by substituting the values into the system.

5r-4s-3t=3 t=s+r r=3s+1

(I), (II), (III): Substitute values

5( -2)-4( -1)-3( -3)? =3 -3? = -1+( -2) -2? =3( -1)+1

(I), (III): Multiply

-10+4+9? =3 -3? =-1+(-2) -2? =-3+1

(I), (II), (III): Add and subtract terms

3=3 ✓ -3=-3 ✓ -2=-2 ✓

Since the substitution of our answers into the given equations resulted in three identities, we know that our solution is correct.