Solve for $X$: $4+2X > 7$
$begingroup$
Solve for $$4+2x>7$$.
My answer was 1 and 1/2 but I am not so sure it is right, could anyone confirm it for me. i need someone to solve the inequality. the first step would be to subtract $4$ from both sides.
algebra-precalculus
edited Jan 20 '14 at 15:35
Yiorgos S. Smyrlis
63k1384163
asked Jan 20 '14 at 15:14
HasanHasan
111
$endgroup$
add a comment |
$begingroup$
Solve for $$4+2x>7$$.
My answer was 1 and 1/2 but I am not so sure it is right, could anyone confirm it for me. i need someone to solve the inequality. the first step would be to subtract $4$ from both sides.
algebra-precalculus
edited Jan 20 '14 at 15:35
Yiorgos S. Smyrlis
63k1384163
asked Jan 20 '14 at 15:14
HasanHasan
111
$endgroup$
1
$begingroup$
It is an inequality, so you cannot have just one answer. You would have to have a spectrum of answers. Yiur answer is correct - just divide the answer that you get by subtracting 4 from each side to get tbe highest bound on the value of x.
$endgroup$
– user122283
Jan 20 '14 at 15:28
$begingroup$
That being said, 1/2 is not correct since $4+2(1/2) = 5 not >7$
$endgroup$
– Squirtle
Jan 20 '14 at 18:05
add a comment |
$begingroup$
Solve for $$4+2x>7$$.
My answer was 1 and 1/2 but I am not so sure it is right, could anyone confirm it for me. i need someone to solve the inequality. the first step would be to subtract $4$ from both sides.
algebra-precalculus
edited Jan 20 '14 at 15:35
Yiorgos S. Smyrlis
63k1384163
asked Jan 20 '14 at 15:14
HasanHasan
111
$endgroup$
Solve for $$4+2x>7$$.
My answer was 1 and 1/2 but I am not so sure it is right, could anyone confirm it for me. i need someone to solve the inequality. the first step would be to subtract $4$ from both sides.
algebra-precalculus
algebra-precalculus
edited Jan 20 '14 at 15:35
Yiorgos S. Smyrlis
63k1384163
asked Jan 20 '14 at 15:14
HasanHasan
111
edited Jan 20 '14 at 15:35
Yiorgos S. Smyrlis
63k1384163
asked Jan 20 '14 at 15:14
HasanHasan
111
edited Jan 20 '14 at 15:35
Yiorgos S. Smyrlis
63k1384163
edited Jan 20 '14 at 15:35
Yiorgos S. Smyrlis
63k1384163
edited Jan 20 '14 at 15:35
Yiorgos S. Smyrlis
63k1384163
63k1384163
asked Jan 20 '14 at 15:14
HasanHasan
111
asked Jan 20 '14 at 15:14
HasanHasan
111
asked Jan 20 '14 at 15:14
HasanHasan
111
111
1
$begingroup$
It is an inequality, so you cannot have just one answer. You would have to have a spectrum of answers. Yiur answer is correct - just divide the answer that you get by subtracting 4 from each side to get tbe highest bound on the value of x.
$endgroup$
– user122283
Jan 20 '14 at 15:28
$begingroup$
That being said, 1/2 is not correct since $4+2(1/2) = 5 not >7$
$endgroup$
– Squirtle
Jan 20 '14 at 18:05
add a comment |
1
$begingroup$
It is an inequality, so you cannot have just one answer. You would have to have a spectrum of answers. Yiur answer is correct - just divide the answer that you get by subtracting 4 from each side to get tbe highest bound on the value of x.
$endgroup$
– user122283
Jan 20 '14 at 15:28
$begingroup$
That being said, 1/2 is not correct since $4+2(1/2) = 5 not >7$
$endgroup$
– Squirtle
Jan 20 '14 at 18:05
1
1
$begingroup$
It is an inequality, so you cannot have just one answer. You would have to have a spectrum of answers. Yiur answer is correct - just divide the answer that you get by subtracting 4 from each side to get tbe highest bound on the value of x.
$endgroup$
– user122283
Jan 20 '14 at 15:28
$begingroup$
It is an inequality, so you cannot have just one answer. You would have to have a spectrum of answers. Yiur answer is correct - just divide the answer that you get by subtracting 4 from each side to get tbe highest bound on the value of x.
$endgroup$
– user122283
Jan 20 '14 at 15:28
$begingroup$
That being said, 1/2 is not correct since $4+2(1/2) = 5 not >7$
$endgroup$
– Squirtle
Jan 20 '14 at 18:05
$begingroup$
That being said, 1/2 is not correct since $4+2(1/2) = 5 not >7$
$endgroup$
– Squirtle
Jan 20 '14 at 18:05
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Yes. You'll have
$$begin{align}4+2xgt 7&iff 4+2x-4gt 7-4\&iff 2xgt 3\&iff 2x/2gt 3/2 (because 2gt 0)\&iff xgt 3/2=1.5.end{align}$$
Note that if $agt 0$, then
$$bgt ciff b/agt c/a.$$
answered Jan 20 '14 at 15:16
mathlovemathlove
91.7k881215
$endgroup$
add a comment |
$begingroup$
Property $bf1$: If $x,y$ are two real numbers such that: $x>y$, we have that $x+mathrm c>y+mathrm c$ for every $mathrm c$ in $Bbb R$.
Property $bf2$: If $x,y$ are two real numbers such that: $x>y$, we have that $xcdotmathrm c>ycdotmathrm c$ for every positive $mathrm c$ in $Bbb R$.
$$require{cancel}begin{align}
4+2xgt 7&overset{mathbf{Pr. 1}}{iff} 4+2xcolor{red}{-4}gt 7color{red}{-4}\,\
& iff 2x gt 3 \ ,\
& overset{mathbf{Pr. 2}}{iff}2xcdotdfrac12 gt 3dfrac 12 \ , \
& iff xgt 3/2=1.5
end{align}$$
$endgroup$
add a comment |
$begingroup$
Yes, after subtracting $ 4 $ to both sides
$$ 2x > 3 $$
Then divide $ 2 $ to both sides to solve for x:
$$ x > frac{3}{2} $$ or $$ x > 1.5 $$
answered Jan 8 at 3:57
Marvin CohenMarvin Cohen
112117
$endgroup$
add a comment |
Your Answer
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3 Answers
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active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes. You'll have
$$begin{align}4+2xgt 7&iff 4+2x-4gt 7-4\&iff 2xgt 3\&iff 2x/2gt 3/2 (because 2gt 0)\&iff xgt 3/2=1.5.end{align}$$
Note that if $agt 0$, then
$$bgt ciff b/agt c/a.$$
answered Jan 20 '14 at 15:16
mathlovemathlove
91.7k881215
$endgroup$
add a comment |
$begingroup$
Yes. You'll have
$$begin{align}4+2xgt 7&iff 4+2x-4gt 7-4\&iff 2xgt 3\&iff 2x/2gt 3/2 (because 2gt 0)\&iff xgt 3/2=1.5.end{align}$$
Note that if $agt 0$, then
$$bgt ciff b/agt c/a.$$
answered Jan 20 '14 at 15:16
mathlovemathlove
91.7k881215
$endgroup$
add a comment |
$begingroup$
Yes. You'll have
$$begin{align}4+2xgt 7&iff 4+2x-4gt 7-4\&iff 2xgt 3\&iff 2x/2gt 3/2 (because 2gt 0)\&iff xgt 3/2=1.5.end{align}$$
Note that if $agt 0$, then
$$bgt ciff b/agt c/a.$$
answered Jan 20 '14 at 15:16
mathlovemathlove
91.7k881215
$endgroup$
Yes. You'll have
$$begin{align}4+2xgt 7&iff 4+2x-4gt 7-4\&iff 2xgt 3\&iff 2x/2gt 3/2 (because 2gt 0)\&iff xgt 3/2=1.5.end{align}$$
Note that if $agt 0$, then
$$bgt ciff b/agt c/a.$$
answered Jan 20 '14 at 15:16
mathlovemathlove
91.7k881215
edited Jan 20 '14 at 15:23
answered Jan 20 '14 at 15:16
mathlovemathlove
91.7k881215
answered Jan 20 '14 at 15:16
mathlovemathlove
91.7k881215
answered Jan 20 '14 at 15:16
mathlovemathlove
91.7k881215
91.7k881215
add a comment |
add a comment |
$begingroup$
Property $bf1$: If $x,y$ are two real numbers such that: $x>y$, we have that $x+mathrm c>y+mathrm c$ for every $mathrm c$ in $Bbb R$.
Property $bf2$: If $x,y$ are two real numbers such that: $x>y$, we have that $xcdotmathrm c>ycdotmathrm c$ for every positive $mathrm c$ in $Bbb R$.
$$require{cancel}begin{align}
4+2xgt 7&overset{mathbf{Pr. 1}}{iff} 4+2xcolor{red}{-4}gt 7color{red}{-4}\,\
& iff 2x gt 3 \ ,\
& overset{mathbf{Pr. 2}}{iff}2xcdotdfrac12 gt 3dfrac 12 \ , \
& iff xgt 3/2=1.5
end{align}$$
$endgroup$
add a comment |
$begingroup$
Property $bf1$: If $x,y$ are two real numbers such that: $x>y$, we have that $x+mathrm c>y+mathrm c$ for every $mathrm c$ in $Bbb R$.
Property $bf2$: If $x,y$ are two real numbers such that: $x>y$, we have that $xcdotmathrm c>ycdotmathrm c$ for every positive $mathrm c$ in $Bbb R$.
$$require{cancel}begin{align}
4+2xgt 7&overset{mathbf{Pr. 1}}{iff} 4+2xcolor{red}{-4}gt 7color{red}{-4}\,\
& iff 2x gt 3 \ ,\
& overset{mathbf{Pr. 2}}{iff}2xcdotdfrac12 gt 3dfrac 12 \ , \
& iff xgt 3/2=1.5
end{align}$$
$endgroup$
add a comment |
$begingroup$
Property $bf1$: If $x,y$ are two real numbers such that: $x>y$, we have that $x+mathrm c>y+mathrm c$ for every $mathrm c$ in $Bbb R$.
Property $bf2$: If $x,y$ are two real numbers such that: $x>y$, we have that $xcdotmathrm c>ycdotmathrm c$ for every positive $mathrm c$ in $Bbb R$.
$$require{cancel}begin{align}
4+2xgt 7&overset{mathbf{Pr. 1}}{iff} 4+2xcolor{red}{-4}gt 7color{red}{-4}\,\
& iff 2x gt 3 \ ,\
& overset{mathbf{Pr. 2}}{iff}2xcdotdfrac12 gt 3dfrac 12 \ , \
& iff xgt 3/2=1.5
end{align}$$
$endgroup$
Property $bf1$: If $x,y$ are two real numbers such that: $x>y$, we have that $x+mathrm c>y+mathrm c$ for every $mathrm c$ in $Bbb R$.
Property $bf2$: If $x,y$ are two real numbers such that: $x>y$, we have that $xcdotmathrm c>ycdotmathrm c$ for every positive $mathrm c$ in $Bbb R$.
$$require{cancel}begin{align}
4+2xgt 7&overset{mathbf{Pr. 1}}{iff} 4+2xcolor{red}{-4}gt 7color{red}{-4}\,\
& iff 2x gt 3 \ ,\
& overset{mathbf{Pr. 2}}{iff}2xcdotdfrac12 gt 3dfrac 12 \ , \
& iff xgt 3/2=1.5
end{align}$$
answered Jan 31 '14 at 22:25
user93957
add a comment |
add a comment |
$begingroup$
Yes, after subtracting $ 4 $ to both sides
$$ 2x > 3 $$
Then divide $ 2 $ to both sides to solve for x:
$$ x > frac{3}{2} $$ or $$ x > 1.5 $$
answered Jan 8 at 3:57
Marvin CohenMarvin Cohen
112117
$endgroup$
add a comment |
$begingroup$
Yes, after subtracting $ 4 $ to both sides
$$ 2x > 3 $$
Then divide $ 2 $ to both sides to solve for x:
$$ x > frac{3}{2} $$ or $$ x > 1.5 $$
answered Jan 8 at 3:57
Marvin CohenMarvin Cohen
112117
$endgroup$
add a comment |
$begingroup$
Yes, after subtracting $ 4 $ to both sides
$$ 2x > 3 $$
Then divide $ 2 $ to both sides to solve for x:
$$ x > frac{3}{2} $$ or $$ x > 1.5 $$
answered Jan 8 at 3:57
Marvin CohenMarvin Cohen
112117
$endgroup$
Yes, after subtracting $ 4 $ to both sides
$$ 2x > 3 $$
Then divide $ 2 $ to both sides to solve for x:
$$ x > frac{3}{2} $$ or $$ x > 1.5 $$
answered Jan 8 at 3:57
Marvin CohenMarvin Cohen
112117
answered Jan 8 at 3:57
Marvin CohenMarvin Cohen
112117
answered Jan 8 at 3:57
Marvin CohenMarvin Cohen
112117
answered Jan 8 at 3:57
Marvin CohenMarvin Cohen
112117
112117
add a comment |
add a comment |
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1
$begingroup$
It is an inequality, so you cannot have just one answer. You would have to have a spectrum of answers. Yiur answer is correct - just divide the answer that you get by subtracting 4 from each side to get tbe highest bound on the value of x.
$endgroup$
– user122283
Jan 20 '14 at 15:28
$begingroup$
That being said, 1/2 is not correct since $4+2(1/2) = 5 not >7$
$endgroup$
– Squirtle
Jan 20 '14 at 18:05