How do you solve more difficult simultaneous equations which involve indices?
Answer
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Hint: According to the question, we can try to solve it by substitution method. Simultaneous equations are equations that contain more than one unknown variable and have that many numbers of equations related to each other.
Complete step-by-step answer:
Let us take an example to solve this question. We will try to solve the simultaneous equations for ‘x’ and ‘y’. The equations are \[\dfrac{1}{{\sqrt {x + 6} }} = \dfrac{3}{{\sqrt y }}\] and \[2y + 3x = 3\] .
Now, we will start with our first equation \[\dfrac{1}{{\sqrt {x + 6} }} = \dfrac{3}{{\sqrt y }}\] . We will try to make \[\sqrt y \] alone. We will shift the terms on other side of the equation, and we get:
\[\sqrt y ={3}.\sqrt {x + 6} \]
Here, we will shift the denominator so that the denominator gets flipped and gets multiplied.
\[\sqrt y ={3}.\sqrt {x + 6} \]
Now, we will remove the square root from \[\sqrt y \] , and we get:
\[ \Rightarrow y = {(3\sqrt {x + 6} )^2}\]
\[ \Rightarrow y = 9x + 54\]
Here, we got the value of ‘y’. So, we will put this value of ‘y’ in the second equation, and we get:
\[2(9x + 54) + 3x = 3\]
We will solve this equation, and we get:
\[ \Rightarrow 18x + 108 + 3x = 3\]
\[ \Rightarrow 21x = - 105\]
Now, we will solve for ‘x’, and we get:
\[ \Rightarrow x = \dfrac{{ - 105}}{{21}}\]
\[ \Rightarrow x = - 5\]
Therefore, we got \[x = - 5\]
When we put the value of ‘x’ in the second equation, we get:
\[2y + 3( - 5) = 3\]
Now, we will solve the equation for ‘y’, and we get:
\[ \Rightarrow 2y - 15 = 3\]
\[ \Rightarrow 2y = 18\]
\[ \Rightarrow y = 9\]
Therefore, we got the value of \[x = - 5\] and \[y = 9\] .
Note: According to the above question, there is no such method for solving any simultaneous equations that involves indices. We generally solve this type of problems by substitution method. This is done because the elimination method is generally not possible.
Complete step-by-step answer:
Let us take an example to solve this question. We will try to solve the simultaneous equations for ‘x’ and ‘y’. The equations are \[\dfrac{1}{{\sqrt {x + 6} }} = \dfrac{3}{{\sqrt y }}\] and \[2y + 3x = 3\] .
Now, we will start with our first equation \[\dfrac{1}{{\sqrt {x + 6} }} = \dfrac{3}{{\sqrt y }}\] . We will try to make \[\sqrt y \] alone. We will shift the terms on other side of the equation, and we get:
\[\sqrt y ={3}.\sqrt {x + 6} \]
Here, we will shift the denominator so that the denominator gets flipped and gets multiplied.
\[\sqrt y ={3}.\sqrt {x + 6} \]
Now, we will remove the square root from \[\sqrt y \] , and we get:
\[ \Rightarrow y = {(3\sqrt {x + 6} )^2}\]
\[ \Rightarrow y = 9x + 54\]
Here, we got the value of ‘y’. So, we will put this value of ‘y’ in the second equation, and we get:
\[2(9x + 54) + 3x = 3\]
We will solve this equation, and we get:
\[ \Rightarrow 18x + 108 + 3x = 3\]
\[ \Rightarrow 21x = - 105\]
Now, we will solve for ‘x’, and we get:
\[ \Rightarrow x = \dfrac{{ - 105}}{{21}}\]
\[ \Rightarrow x = - 5\]
Therefore, we got \[x = - 5\]
When we put the value of ‘x’ in the second equation, we get:
\[2y + 3( - 5) = 3\]
Now, we will solve the equation for ‘y’, and we get:
\[ \Rightarrow 2y - 15 = 3\]
\[ \Rightarrow 2y = 18\]
\[ \Rightarrow y = 9\]
Therefore, we got the value of \[x = - 5\] and \[y = 9\] .
Note: According to the above question, there is no such method for solving any simultaneous equations that involves indices. We generally solve this type of problems by substitution method. This is done because the elimination method is generally not possible.
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