Exercise 51 Page 376

To solve a system of linear equations using the Elimination Method, one of the variable terms needs to be eliminated when one equation is added to or subtracted from the other equation. In this exercise, this means that either the q-terms or the r-terms must cancel each other out. q+4 r=- 8 & (I) 3 q+2 r=6 & (II) In its current state, this will not happen. Therefore, we need to find a common multiple between two variable like terms in the system. If we multiply (II) by - 2, the r-terms will have opposite coefficients. q+4 r=- 8 - 2(3 q+2 r)=- 2(6) ⇓ q+ 4r=- 8 - 6 q- 4r=- 12We can see that the r-terms will eliminate each other if we add (I) to (II).

q+4r=- 8 - 6q-4r=- 12

SysEqnAdd

(II): Add (I)

q+4r=- 8 - 6q-4r+( q+4r)=- 12+( - 8)

▼

(II):Solve for q

AddSubTerms

(II): Add and subtract terms

q+4r=- 8 - 5q=- 20

DivEqn

(II): .LHS /(- 5).=.RHS /(- 5).

q+4r=- 8 q=4

Now we can solve for r by substituting the value of q into either equation and simplifying.

q+4r=- 8 q=4

Substitute

(I): q= 4

4+4r=- 8 q=4

▼

(I):Solve for r

SubEqn

(I): LHS-4=RHS-4

4r=- 12 q=4

DivEqn

(I): .LHS /4.=.RHS /4.

r=- 3 q=4

The solution, or point of intersection, of the system of equations is (4,- 3).