
Solve the following simultaneous equation:
$\begin{align}
& 23a+17b=103, \\
& 17a+23b=97. \\
\end{align}$
Answer
508.8k+ views
Hint: We are going to solve the two simultaneous equations by elimination method. Multiply the first equation by 17 and then multiply the second equation by 23. Subtracting the first equation from the second equation will give you the value of b then substitute this value of “b” in the first equation, you will get the value of “a”.
Complete step-by-step answer:
The two simultaneous equations given in the question are:
$\begin{align}
& 23a+17b=103............Eq.(1) \\
& 17a+23b=97..............Eq.(2) \\
\end{align}$
Multiplying eq. (1) by 17 and multiplying eq. (2) by 23 then subtract this multiplied eq. (1) from multiplied eq. (2) we get,
$\begin{matrix}
\left( 17a+23b=97 \right)\times 23 \\
\dfrac{-\left( 23a+17b=103 \right)\times 17}{\left( {{23}^{2}}-{{17}^{2}} \right)b=\left( 2231-1751 \right)} \\
\end{matrix}$
Simplifying the above equation we get,
$240b=480$
Dividing 240 on both the sides we get,
$b=\dfrac{480}{240}=2$
Substituting the above value of b in eq. (1) we get,
$\begin{align}
& 23a+17\left( 2 \right)=103 \\
& \Rightarrow 23a+34=103 \\
& \Rightarrow 23a=103-34 \\
& \Rightarrow 23a=69 \\
& \Rightarrow a=3 \\
\end{align}$
From the above solving equations by elimination method we have found the value of a = 3 and b = 2.
Hence, the value of a = 3 and b = 2 in the given simultaneous equation.
Note: We can also solve the value of “a” & “b” in the given system of simultaneous equations by substitution method.
$\begin{align}
& 23a+17b=103............Eq.(1) \\
& 17a+23b=97..............Eq.(2) \\
\end{align}$
From equation (1), writing “a” in terms of “b” as follows:
$\begin{align}
& 23a=103-17b \\
& \Rightarrow a=\dfrac{103-17b}{23} \\
\end{align}$
Now, plugging this value of “a” in the equation (2) we get,
$\begin{align}
& 17\left( \dfrac{103-17b}{23} \right)+23b=97 \\
& \Rightarrow 17\left( 103 \right)-{{17}^{2}}b+{{23}^{2}}b=97\left( 23 \right) \\
& \Rightarrow 240b=2231-1751 \\
& \Rightarrow 240b=480 \\
& \Rightarrow b=2 \\
\end{align}$
Substituting this value of b in eq. (1) we get,
$\begin{align}
& 23a+17\left( 2 \right)=103 \\
& \Rightarrow 23a+34=103 \\
& \Rightarrow 23a=103-34 \\
& \Rightarrow 23a=69 \\
& \Rightarrow a=3 \\
\end{align}$
From the above calculations, we are getting the value of a = 3 and b = 2 which is the same as that we have got the solution part of the question. So, you can use both the methods as per your convenience.
It’s always better to use the method which will minimize the calculations as it will save your time in calculation and prevent calculation mistakes.
Complete step-by-step answer:
The two simultaneous equations given in the question are:
$\begin{align}
& 23a+17b=103............Eq.(1) \\
& 17a+23b=97..............Eq.(2) \\
\end{align}$
Multiplying eq. (1) by 17 and multiplying eq. (2) by 23 then subtract this multiplied eq. (1) from multiplied eq. (2) we get,
$\begin{matrix}
\left( 17a+23b=97 \right)\times 23 \\
\dfrac{-\left( 23a+17b=103 \right)\times 17}{\left( {{23}^{2}}-{{17}^{2}} \right)b=\left( 2231-1751 \right)} \\
\end{matrix}$
Simplifying the above equation we get,
$240b=480$
Dividing 240 on both the sides we get,
$b=\dfrac{480}{240}=2$
Substituting the above value of b in eq. (1) we get,
$\begin{align}
& 23a+17\left( 2 \right)=103 \\
& \Rightarrow 23a+34=103 \\
& \Rightarrow 23a=103-34 \\
& \Rightarrow 23a=69 \\
& \Rightarrow a=3 \\
\end{align}$
From the above solving equations by elimination method we have found the value of a = 3 and b = 2.
Hence, the value of a = 3 and b = 2 in the given simultaneous equation.
Note: We can also solve the value of “a” & “b” in the given system of simultaneous equations by substitution method.
$\begin{align}
& 23a+17b=103............Eq.(1) \\
& 17a+23b=97..............Eq.(2) \\
\end{align}$
From equation (1), writing “a” in terms of “b” as follows:
$\begin{align}
& 23a=103-17b \\
& \Rightarrow a=\dfrac{103-17b}{23} \\
\end{align}$
Now, plugging this value of “a” in the equation (2) we get,
$\begin{align}
& 17\left( \dfrac{103-17b}{23} \right)+23b=97 \\
& \Rightarrow 17\left( 103 \right)-{{17}^{2}}b+{{23}^{2}}b=97\left( 23 \right) \\
& \Rightarrow 240b=2231-1751 \\
& \Rightarrow 240b=480 \\
& \Rightarrow b=2 \\
\end{align}$
Substituting this value of b in eq. (1) we get,
$\begin{align}
& 23a+17\left( 2 \right)=103 \\
& \Rightarrow 23a+34=103 \\
& \Rightarrow 23a=103-34 \\
& \Rightarrow 23a=69 \\
& \Rightarrow a=3 \\
\end{align}$
From the above calculations, we are getting the value of a = 3 and b = 2 which is the same as that we have got the solution part of the question. So, you can use both the methods as per your convenience.
It’s always better to use the method which will minimize the calculations as it will save your time in calculation and prevent calculation mistakes.
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