Sorry, it’s a typo. Challenge- Math(3/5 Difficulty) Solve [tex]\sqrt{2x+3} -\sqrt{2x-5} =\dfrac{8}{7}[/tex]. *Note it’s subtraction instead of addition. Solution is [tex]x=\dfrac{2661}{392}[/tex]. About the author Lydia
Given equation: [tex]\sqrt{2x+3} -\sqrt{2x-5} =\dfrac{8}{7}[/tex] …[1] Solution Consider [tex]\sqrt{2x+3} +\sqrt{2x-5} =k[/tex]. …[2] Multiplying two equations we get, [tex](2x+3)-(2x-5)=\dfrac{8k}{7}[/tex] [tex]\implies 3+5=\dfrac{8k}{7}[/tex] [tex]\implies \dfrac{8k}{7} =8[/tex] [tex]\implies k=7[/tex] Therefore, we have a system equation, [tex]\begin{cases} & \sqrt{2x+3} +\sqrt{2x-5} =7 \\ & \sqrt{2x+3} -\sqrt{2x-5} =\dfrac{8}{7} \end{cases}[/tex] And therefore, [tex]\begin{cases} & \sqrt{2x+3} = \dfrac{57}{14} \\ & \sqrt{2x-5} = \dfrac{41}{14}\end{cases}[/tex] Solving one of the equation, we get [tex]x=\dfrac{2661}{392}[/tex] ✓ Required answer. Reply
[tex]\texttt{\textsf{\large{\underline{Solution}:}}}[/tex] We have to find out x. Given, [tex]\tt\implies \sqrt{2x+3}-\sqrt{2x-5}=\dfrac{8}{7}[/tex] Multiplying both sides by 7, we get, [tex]\tt\implies7(\sqrt{2x+3}-\sqrt{2x-5})=8[/tex] [tex]\tt\implies 7\sqrt{2x+3}-7\sqrt{2x-5}=8[/tex] [tex]\tt\implies 7\sqrt{2x+3}=7\sqrt{2x-5}+8[/tex] Squaring both sides, we get, [tex]\tt\implies (7\sqrt{2x+3})^{2}=(7\sqrt{2x-5}+8)^{2}[/tex] [tex]\tt\implies 49(2x+3)=(7\sqrt{2x-5})^{2}+(8)^{2} + 2\times8\times(7\sqrt{2x-5})[/tex] [tex]\tt\implies 49(2x+3)=49(2x-5)+64 + 112\sqrt{2x-5}[/tex] [tex]\tt\implies 98x+147=98x-245+64 + 112\sqrt{2x-5}[/tex] Cancel out like terms. We get, [tex]\tt\implies 147=-245+64 + 112\sqrt{2x-5}[/tex] [tex]\tt\implies 147+245-64=112\sqrt{2x-5}[/tex] [tex]\tt\implies 328=112\sqrt{2x-5}[/tex] [tex]\tt\implies \sqrt{2x-5}=\dfrac{328}{112}[/tex] [tex]\tt\implies \sqrt{2x-5}=\dfrac{41}{14}[/tex] Squaring both sides, we get, [tex]\tt\implies 2x-5=\dfrac{1681}{196}[/tex] Adding 5 from both sides, we get, [tex]\tt\implies 2x=\dfrac{1681}{196}+5[/tex] [tex]\tt\implies 2x=\dfrac{1681+5\times196}{196}[/tex] [tex]\tt\implies 2x=\dfrac{1681+980}{196}[/tex] [tex]\tt\implies 2x=\dfrac{2661}{196}[/tex] Dividing both sides by 2, we get, [tex]\tt\implies x=\dfrac{2661}{392}[/tex] ⊕ This is the required answer for the question. [tex]\texttt{\textsf{\large{\underline{Answer}:}}}[/tex] x = 2661/392 Reply
Given equation: [tex]\sqrt{2x+3} -\sqrt{2x-5} =\dfrac{8}{7}[/tex] …[1]
Solution
Consider [tex]\sqrt{2x+3} +\sqrt{2x-5} =k[/tex]. …[2]
Multiplying two equations we get,
[tex](2x+3)-(2x-5)=\dfrac{8k}{7}[/tex]
[tex]\implies 3+5=\dfrac{8k}{7}[/tex]
[tex]\implies \dfrac{8k}{7} =8[/tex]
[tex]\implies k=7[/tex]
Therefore, we have a system equation,
[tex]\begin{cases} & \sqrt{2x+3} +\sqrt{2x-5} =7 \\ & \sqrt{2x+3} -\sqrt{2x-5} =\dfrac{8}{7} \end{cases}[/tex]
And therefore,
[tex]\begin{cases} & \sqrt{2x+3} = \dfrac{57}{14} \\ & \sqrt{2x-5} = \dfrac{41}{14}\end{cases}[/tex]
Solving one of the equation, we get
[tex]x=\dfrac{2661}{392}[/tex]
✓ Required answer.
[tex]\texttt{\textsf{\large{\underline{Solution}:}}}[/tex]
We have to find out x.
Given,
[tex]\tt\implies \sqrt{2x+3}-\sqrt{2x-5}=\dfrac{8}{7}[/tex]
Multiplying both sides by 7, we get,
[tex]\tt\implies7(\sqrt{2x+3}-\sqrt{2x-5})=8[/tex]
[tex]\tt\implies 7\sqrt{2x+3}-7\sqrt{2x-5}=8[/tex]
[tex]\tt\implies 7\sqrt{2x+3}=7\sqrt{2x-5}+8[/tex]
Squaring both sides, we get,
[tex]\tt\implies (7\sqrt{2x+3})^{2}=(7\sqrt{2x-5}+8)^{2}[/tex]
[tex]\tt\implies 49(2x+3)=(7\sqrt{2x-5})^{2}+(8)^{2} + 2\times8\times(7\sqrt{2x-5})[/tex]
[tex]\tt\implies 49(2x+3)=49(2x-5)+64 + 112\sqrt{2x-5}[/tex]
[tex]\tt\implies 98x+147=98x-245+64 + 112\sqrt{2x-5}[/tex]
Cancel out like terms. We get,
[tex]\tt\implies 147=-245+64 + 112\sqrt{2x-5}[/tex]
[tex]\tt\implies 147+245-64=112\sqrt{2x-5}[/tex]
[tex]\tt\implies 328=112\sqrt{2x-5}[/tex]
[tex]\tt\implies \sqrt{2x-5}=\dfrac{328}{112}[/tex]
[tex]\tt\implies \sqrt{2x-5}=\dfrac{41}{14}[/tex]
Squaring both sides, we get,
[tex]\tt\implies 2x-5=\dfrac{1681}{196}[/tex]
Adding 5 from both sides, we get,
[tex]\tt\implies 2x=\dfrac{1681}{196}+5[/tex]
[tex]\tt\implies 2x=\dfrac{1681+5\times196}{196}[/tex]
[tex]\tt\implies 2x=\dfrac{1681+980}{196}[/tex]
[tex]\tt\implies 2x=\dfrac{2661}{196}[/tex]
Dividing both sides by 2, we get,
[tex]\tt\implies x=\dfrac{2661}{392}[/tex]
⊕ This is the required answer for the question.
[tex]\texttt{\textsf{\large{\underline{Answer}:}}}[/tex]