Exercise 1 Page 218

Does substituting the point into every equation in the system result in true statements?

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To determine if a point satisfies an equation, we substitute the point into the equation and simplify. If the resulting statement is true, the point is a solution of the equation. For systems of equations, we can use the same method. However, substituting the point must create true statements in every equation in the system. Let's test the given point.

2x+y=0 & (I) - x+2y=5 & (II)

SubstituteII

x= 1, y= - 2

2( 1)+( - 2)? =0 - 1+2( - 2)? =5

IdPropMult

(I): Identity Property of Multiplication

2+(- 2)? =0 - 1+2(- 2)? =5

AddNeg

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

2-2? =0 - 1+2(- 2)? =5

MultPosNeg

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

2-2? =0 - 1-4? =5

(I), (II): Subtract term

0=0 ✓ - 5≠ 5 *

Since - 5 is not equal to 5, the statement is not true after substituting the point. Therefore, the point (1,- 2) satisfies (I) but does not satisfy (II). Because one of the statements is not true, the point (1,- 2) is not a solution of the system.

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