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DIRECTIONS :
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- Find the solution/s of the following mathematical sentences. Describe these mathematical sentences.
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SOLUTION :
1) b+3 < 10
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- Subtract 3 from both sides to get b < 10 - 3 then subtract 3 from 10 to have 7,so therefore the answer is [tex]\tt\red{\underline{ \: b < 7}}[/tex]
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2) 4c - 5 > 19
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- First, let's add 5 to both sides,4c > 19 + 5, then add 19 + 5 to get 24,4c > 24.divide both sides by 4,since 4 is positive, the inequality direction remains the same , [tex]c>\frac{24}{4}[/tex] , then divide 24 by 4 to get 6,[tex] \tt\red{\underline{ \: c>6 }}[/tex]
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3) 13 - 7m > -1
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- Subtract 13 from both sides, -7m > -1 -13, subtract 13 from -1 to get -14, -7m > -14,then divide both sides by -7, since -7 is negative, the inequality direction changed, [tex]m<\frac{-14}{-7},[/tex] divide -14 by -7 to get 2, [tex]\tt\red{\underline{ \: m<2 }}[/tex]
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4) r² + 6r - 7 = 0
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FORMULA :
- r² + ( a + b) r + ab = ( r + a ) ( r + b )
SOLUTION :
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now let's set up a system to be solved :
- [tex]\tt{a+b=6, ab=-7 }[/tex]
- [tex]\tt{a=-1,b=7 }[/tex]
- [tex]\tt{\left(r-1\right)\left(r+7\right) }[/tex]
- [tex] \tt\red{{ \:r=1 \: or \:r=-7 }}[/tex]
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Therefore,the answer is [tex] \tt\red{\underline{ \:r=1,r=-7 }}[/tex]
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5) 2m² = 50
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- Divide both sides by 2,[tex]m^{2}=\frac{50}{2} [/tex], then divide 50 by 2 to get 25,m² = 25, take the square root of both sides of the equation to get the answer, [tex]\tt\red{\underline{ \: m=5 \:or \:m=-5 }}[/tex]
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6) n² - 7n = - 10
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FORMULA :
- n² + ( a + b) n + ab = ( n + a ) ( n + b )
SOLUTION :
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now let's set up a system to be solved :
- [tex]\tt{ a+b=-7 \:,\: ab=10 }[/tex]
- [tex]\tt{ -1,-10\:, \:-2,-5 }[/tex]
- [tex]\tt{ -1-10=-11 \:,\:-2-5=-7 }[/tex]
- [tex]\tt{ a=-5\:,\:b=-2 }[/tex]
- [tex]\tt\red{ \left(n-5\right)\:,\:\left(n-2\right) }[/tex]
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Therefore, the answer is [tex]\tt\red{\underline{ \: n=5 \:,\:n=2 }}[/tex]
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[tex]\tiny\begin{array}{|c|c|c|}\hline \\ \sf \: Quadratic \: Equations & \sf \: Quadratic \: Inequalities & \sf \: Not \: a \: Quadratic \: Equation \: nor \: a \: Quadratic \: Inequalities \\ \hline \\ \red{5 + 6k = 7 {k}^{2} } \: & \red{3 {x}^{2} + x + 10 > 0 } & \red{m + 5 <_ { - 12} } \\ \hline \\ \red{ {x}^{2} - 4x + 8 = 0} \: & \red{( \: 2 {n}^{2} + 5 \: )( \: 3n - 1 \: ) } & \red{7 {d}^{2} < _ { 14} } \\ \hline \\ \: & \red{2 {s}^{2} - 3s + 5 < 0 } & \red{(2r - 3)(r + 1) < _ { 2} } \\ \hline \end{array}[/tex]
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#CarryOnLearning
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