HAPTER 1
nequalities
Solve 3 + 2*<7.
Answer x<2 [Divide both sides by 2. This is equivalent to multiplying by 5.] In interval notation, the
solution is the set (—°°, 2).
Solve 5 - 3* < 5x + 2.
Answer 1
Answer — 6 < x < 2 [Divide by 2.] In interval notation, the solution is the set (—6,2).
Solve 3<4x-l<5.
Answer 1 s x < \ [Divide by 4.] In interval notation, the solution is the set [1, |).
Solve 4<-2x + 5<7.
Answer \ >*>-! [Divide by -2. Since -2 is negative, we must reverse the inequalities.] In interval
notation, the solution is the set [-1, |).
Solve 5 < \x. + 1 s 6.
Answer 12<^sl5 [Multiply by 3.] In interval notation, the solution is the set [12,15].
Solve 2/jc<3.
Case 2. x<0. 2/x<3. 2>3x [Multiply by jr. Reverse the inequality.], |>jc [Divide by 3.] Notice
that this condition |>x is satisfied whenever jc<0. Hence, in the case where x<0, the inequality is
satisfied by all such x.
Answer f < x or x < 0. As shown in Fig. 1-1, the solution is the union of the intervals (1,«) and (—°°, 0).
Solve
negative. Case 1. x-3>0 [This is equivalent to x>3.] Multiplying the given inequality (1) by the
positive quantity x-3 preserves the inequality: * + 4<2;t-6, 4
is equivalent to x<3]. Multiplying the given inequality (1) by the negative quantity x — 3 reverses the
inequality: * + 4>2*-6, 4>x-6 [Subtract*.], 10>x [Add 6.] Thus, when x<3, the inequality
1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
Fig. 1-1
2x < 4 [Subtract 3 from both sides. This is equivalent to adding -3 to both sides.]
5-3>x<5x + 2, 5<8* + 2 [Add 3x to both sides.], 3<8* [Subtract 2 from both sides.]
-7 < 2* + 5 < 9, -12 < 2x < 4 [Subtract 5 from all terms.]
3<4x-l<5, 4<4x<6 [Add 1 to all terms.]
4<-2x + 5<7, -K-2jc<2 [Subtracts.]
5<|x + l<6, 4<|*s5 [Subtract 1.]
x may be positive or negative. Case 1. x>0. 2/x<3. 2<3x [Multiply by AC.], |
nequalities
Solve 3 + 2*<7.
Answer x<2 [Divide both sides by 2. This is equivalent to multiplying by 5.] In interval notation, the
solution is the set (—°°, 2).
Solve 5 - 3* < 5x + 2.
Answer 1
Answer — 6 < x < 2 [Divide by 2.] In interval notation, the solution is the set (—6,2).
Solve 3<4x-l<5.
Answer 1 s x < \ [Divide by 4.] In interval notation, the solution is the set [1, |).
Solve 4<-2x + 5<7.
Answer \ >*>-! [Divide by -2. Since -2 is negative, we must reverse the inequalities.] In interval
notation, the solution is the set [-1, |).
Solve 5 < \x. + 1 s 6.
Answer 12<^sl5 [Multiply by 3.] In interval notation, the solution is the set [12,15].
Solve 2/jc<3.
Case 2. x<0. 2/x<3. 2>3x [Multiply by jr. Reverse the inequality.], |>jc [Divide by 3.] Notice
that this condition |>x is satisfied whenever jc<0. Hence, in the case where x<0, the inequality is
satisfied by all such x.
Answer f < x or x < 0. As shown in Fig. 1-1, the solution is the union of the intervals (1,«) and (—°°, 0).
Solve
negative. Case 1. x-3>0 [This is equivalent to x>3.] Multiplying the given inequality (1) by the
positive quantity x-3 preserves the inequality: * + 4<2;t-6, 4
is equivalent to x<3]. Multiplying the given inequality (1) by the negative quantity x — 3 reverses the
inequality: * + 4>2*-6, 4>x-6 [Subtract*.], 10>x [Add 6.] Thus, when x<3, the inequality
1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
Fig. 1-1
2x < 4 [Subtract 3 from both sides. This is equivalent to adding -3 to both sides.]
5-3>x<5x + 2, 5<8* + 2 [Add 3x to both sides.], 3<8* [Subtract 2 from both sides.]
-7 < 2* + 5 < 9, -12 < 2x < 4 [Subtract 5 from all terms.]
3<4x-l<5, 4<4x<6 [Add 1 to all terms.]
4<-2x + 5<7, -K-2jc<2 [Subtracts.]
5<|x + l<6, 4<|*s5 [Subtract 1.]
x may be positive or negative. Case 1. x>0. 2/x<3. 2<3x [Multiply by AC.], |