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using substitution, identify if the given point is a solution of the linear inequality. write TRUE if the point is a solution of the inequality and FALSE if not
_1.5x - y > -2;(-3,-4)
_2.4x - y < -8;(-1,-10)
_3.y > 3x; (0,0)
_4.x + 2y < 10; (5,-3)

_6.y > 3x + 1; (-1,1)


Sagot :

• Problem:

Using substitution, identify if the given point is a solution of the linear inequality. Write TRUE if the point is a solution of the inequality and FALSE if not.

• Answers:

1. 5x - y > -2; (-3,-4)

[tex] \large \boxed{ \begin{array}{} \tt 5x - y > -2; (-3,-4) \\ \tt5( - 3) - ( - 4) > - 2 \\ \tt - 15 + 4 > - 2 \\ \tt - 11 > - 2 \\ \implies \tt false\end{array}}[/tex]

2. 4x - y < -8; (-1,-10)

[tex] \large \boxed{ \begin{array}{} \tt 4x - y < -8; (-1,-10) \\ \tt4( - 1) - ( - 10) < - 8 \\ \tt - 4 + 10 < - 8 \\ \tt6 < - 8 \\ \tt \implies false\end{array}}[/tex]

3. y > 3x; (0,0)

[tex]\large \boxed{ \begin{array}{} \tt y > 3x; (0,0) \\ \tt0 > 3(0) \\ \tt0 > 0 \\ \tt \implies false\end{array}}[/tex]

4. x + 2y < 10; (5,-3)

[tex]\large \boxed{ \begin{array}{} \tt x + 2y < 10; (5,-3) \\ \tt5 + 2( - 3) < 10 \\ \tt5 - 6 < 10 \\ \tt - 1 < 10 \\ \implies \tt true\end{array}}[/tex]

5. y > 3x + 1; (-1,1)

[tex]\large \boxed{ \begin{array}{} \tt y > 3x + 1; (-1,1) \\ \tt1 > 3( - 1) + 1 \\ \tt1 > - 3 + 1 \\ \tt1 > - 2 \\ \tt \implies true\end{array}}[/tex]