Question

Find those values of $c$ for which the given equations are consistent.

$$2x + 3y = 3 \\\ \left( {c + 2} \right)x + \left( {c + 4} \right)y = c + 6 \\\ {\left( {c + 2} \right)^2}x + {\left( {c + 4} \right)^2}y = {\left( {c + 6} \right)^2} \\\$$

Also solve the above equations for these values of $c$.

Answer 1

Hint : Here, in the question, we have been given a system of linear equations having two variables and one unknown constant $c$. And we are asked to find the values of $c$ for which the given system is consistent and also the solution of the equations. We will use the determinant method to find the values of $c$ and then solve the equation by elimination method.

Complete step-by-step answer :
Given system of equations:

$$2x + 3y = 3 \\\ \left( {c + 2} \right)x + \left( {c + 4} \right)y = c + 6 \\\ {\left( {c + 2} \right)^2}x + {\left( {c + 4} \right)^2}y = {\left( {c + 6} \right)^2} \\\$$

Now, we know for any system of equations to be consistent, the first condition is that the determinant of the coefficients and coefficients should be equal to zero. Therefore,

2&3&3 \\\ {c + 2}&{c + 4}&{c + 6} \\\ {{{\left( {c + 2} \right)}^2}}&{{{\left( {c + 4} \right)}^2}}&{{{\left( {c + 6} \right)}^2}} \end{array}} \right| = 0$$ Applying $${C_1} \to {C_1} - {C_3}$$ and $${C_2} \to {C_2} - {C_3}$$, we get $$\left| {\begin{array}{*{20}{c}} { - 1}&0&3 \\\ { - 4}&{ - 2}&{c + 6} \\\ { - 8c - 32}&{ - 4c - 20}&{{{\left( {c + 6} \right)}^2}} \end{array}} \right| = 0$$ Now, we know if we multiply all elements of a row or a column by any constant $$p$$, it results $$p$$ times in value of the determinant. Therefore, Applying $${C_1} \to \left( { - 1} \right){C_1}$$ and $${C_2} \to \left( { - 1} \right){C_2}$$ to determinant, we get, $$\left| {\begin{array}{*{20}{c}} 1&0&3 \\\ 4&2&{c + 6} \\\ {8c + 32}&{4c + 20}&{{{\left( {c + 6} \right)}^2}} \end{array}} \right| = 0$$ Applying $${C_1} \to {C_1} - 2{C_2}$$, we get $$\left| {\begin{array}{*{20}{c}} 1&0&3 \\\ 0&2&{c + 6} \\\ { - 8}&{4c + 20}&{{{\left( {c + 6} \right)}^2}} \end{array}} \right| = 0$$ Applying $${C_3} \to {C_3} - 3{C_1}$$, we get

\left| {\begin{array}{*{20}{c}}
1&0&0 \\
0&2&{c + 6} \\
{ - 8}&{4c + 20}&{{{\left( {c + 6} \right)}^2} + 24}
\end{array}} \right| = 0 \\
\Rightarrow 1\left[ {2\left( {{{\left( {c + 6} \right)}^2} + 24} \right) - \left( {c + 6} \right)\left( {4c + 20} \right)} \right] - 0 + 0 = 0 \\

$$Simplifying it, we get$$

2\left[ {\left( {{c^2} + 36 + 12c} \right) + 24} \right] - \left( {4{c^2} + 20c + 24c + 120} \right) = 0 \\
\Rightarrow 2{c^2} + 72 + 24c + 48 - 4{c^2} - 44c - 120 = 0 \\
\Rightarrow - 2{c^2} - 20c = 0 \\

Dividing by $$\left( { - 2} \right)$$ both sides, we get

{c^2} + 10c = 0 \\
\Rightarrow c\left( {c + 10} \right) = 0 \\
\Rightarrow c = 0;or;c = - 10 ;

Now, when $$c = 0$$, the system of given equations become:

2x + 3y = 3 \\
2x + 4y = 6 \\
4x + 16y = 36 \\

$$Using elimination method,$$

2x + ;3y = ;;3 \\
2x + ;4y = ;;6 \\
\underline { - ;;; - ;;;;;;; - ;;;} \\
;;;;;; - y = - 3 \\
\Rightarrow y = 3 \\
\Rightarrow x = - 3 ;

Similarly when $$c = - 10$$, the given system of equations become:

2x + 3y = 3 \\
- 8x - 6y = - 4; \Rightarrow 4x + 3y = 2 \\
64x + 36y = 16; \Rightarrow 16x + 9y = 4 ;

$$Using elimination method, we get$$

2x + ;3y = ;;3 \\
4x + 3y = ;;2 \\
\underline { - ;;; - ;;;;;;; - ;} \\
- 2x;;;;;; = ;;1 \\
\Rightarrow x = - \dfrac{1}{2} \\
\Rightarrow y = \dfrac{4}{3} ;

$$**Note** : The system of equations is consistent if the lines are intersecting to each other or are coinciding with each other. In other words, if the system of equations is consistent, the lines cannot be parallel to each other. The solution of a linear equation having two variables can be found using three different methods: substitution method, elimination method and completing square method.$$