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Hint: The method of solving "by substitution" works by solving one of the equations (you choose which one) for one of the variables (you choose which one), and then plugging this back into the other equation, "substituting" for the chosen variable and solving for the other. Then you back-solve the first variable.

We have been given two equation \[2x+3y=9,3x+4y=5\] which we will have to solve by substitution method to solve the equation which is shown below;

\[\begin{align}

& 2x+3y=9.............................\text{(1)} \\

& 3x+4y=5..............................(2) \\

& \text{solving equation (1) we get,} \\

& \text{y=}\dfrac{\text{9-2x}}{3} \\

\end{align}\]

Now, we will substitute the above value of “y” using the equation (1) into equation (2) and then we will solve further for “x” as shown below,

\[\begin{align}

& \Rightarrow 3x+4\left( \dfrac{9-2x}{3} \right)=5 \\

& \Rightarrow \dfrac{9x+36-8x}{3}=5 \\

& \Rightarrow x+36=5\times 3 \\

& \Rightarrow x=15-36 \\

& \Rightarrow x=-21 \\

\end{align}\]

So, on solving the above equation we get x= -21.

Now, we will use the value of “x” in either of the two equation and solve for the value of “y” and we get ,

\[\begin{align}

& \Rightarrow 2x+3y=9 \\

& \Rightarrow 2\times (-21)+3y=9 \\

& \Rightarrow 3y=9+42 \\

& \Rightarrow y=\dfrac{51}{3}=17 \\

\end{align}\]

Hence, we get x= -21 and y=17 as a solution for the given system of equations by using a method of substitution.

NOTE: Remember the method of substitution , in general it is useful for the school examination purpose. Also be careful while doing calculation because there is a chance that you might make a mistake while substituting the value of one variable to the other equation and you will get the incorrect answer.

__Complete step-by-step solution -__We have been given two equation \[2x+3y=9,3x+4y=5\] which we will have to solve by substitution method to solve the equation which is shown below;

\[\begin{align}

& 2x+3y=9.............................\text{(1)} \\

& 3x+4y=5..............................(2) \\

& \text{solving equation (1) we get,} \\

& \text{y=}\dfrac{\text{9-2x}}{3} \\

\end{align}\]

Now, we will substitute the above value of “y” using the equation (1) into equation (2) and then we will solve further for “x” as shown below,

\[\begin{align}

& \Rightarrow 3x+4\left( \dfrac{9-2x}{3} \right)=5 \\

& \Rightarrow \dfrac{9x+36-8x}{3}=5 \\

& \Rightarrow x+36=5\times 3 \\

& \Rightarrow x=15-36 \\

& \Rightarrow x=-21 \\

\end{align}\]

So, on solving the above equation we get x= -21.

Now, we will use the value of “x” in either of the two equation and solve for the value of “y” and we get ,

\[\begin{align}

& \Rightarrow 2x+3y=9 \\

& \Rightarrow 2\times (-21)+3y=9 \\

& \Rightarrow 3y=9+42 \\

& \Rightarrow y=\dfrac{51}{3}=17 \\

\end{align}\]

Hence, we get x= -21 and y=17 as a solution for the given system of equations by using a method of substitution.

NOTE: Remember the method of substitution , in general it is useful for the school examination purpose. Also be careful while doing calculation because there is a chance that you might make a mistake while substituting the value of one variable to the other equation and you will get the incorrect answer.

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