Exercise 26 Page 482

Practice makes perfect

a We are given three line equations, and we know that the triangle is enclosed by them. To find the vertices of this triangle, we need to recall that in this case each vertex is a point of intersection of two of the three given lines. Let's start with evaluating the point of intersection for the first two equations.

x-y=-1 & (I) y=2 & (II) To do this let's use the Substitution Method and substitute the value of y from the second equation into the first one.

x-y=-1 y=2

Substitute

(I):y= 2

x- 2=-1 y=2

AddEqn

(I):LHS+2=RHS+2

x=1 y=2

The first vertex is (1,2). Now, let's find the second vertex and use the two last equations. y=2 & (I) -0.4x-y=-5.2 & (II) We will again use the Substitution Method and substitute the value of y from the first equation into the second one.

y=2 -0.4x-y=-5.2

Substitute

(II):y= 2

y=2 -0.4x- 2=-5.2

AddEqn

(II):LHS+2=RHS+2

y=2 -0.4x=-3.2

DivEqn

(III):.LHS /(-0.4).=.RHS /(-0.4).

y=2 x=8

The second vertex has coordinates of (8,2). Finally, let's find the third vertex using the first and the last equations. x-y=-1 & (I) -0.4x-y=-5.2 & (II) This time we will use the Elimination Method and subtract the second equation from the first one.

x-y=-1 -0.4x-y=-5.2

SysEqnSub

(I):Subtract (II)

x-y-( -0.4x-y)=-1-( -5.2) -0.4x-y=-5.2

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Solve for x

SubNeg

(I):a-(- b)=a+b

x-y+0.4x+y=-1+5.2 -0.4x-y=-5.2

AddTerms

(I):Add terms

1.4x=4.2 -0.4x-y=-5.2