
Find the value of k, infinitely many solution $$2x + 3y = 7,(k - 1)x + (k + 2)y = 3k$$
Answer
563.4k+ views
Hint: We know the general equations i.e.
$ {{a_1}x + b{_1}y = {c_1}} $
$ {{a_2}x + b{_2}y = {c_2}} $
After comparing general equations with the given equation we got the value. After that we need to apply the condition of infinite solution which is $$\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$$ . after this we can get the value of k.
Complete step-by-step answer:
Consider the given equations.
2x + 3y = 7
(k - 1)x + (k + 2)y = 3k
The general equations
$ {{a_1}x + b{_1}y = {c_1}}$
$ {{a_2}x + b{_2}y = {c_2}}$
So,
$ {{a_1} = 2,{b_1} = 3,{c_1} = 7}$
${a_2}$ = k - 1,${b_2}$ = k + 2,${c_2}$ = 3k
We know that the condition of infinite solution
$$\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$$
Therefore,
${\dfrac{2}{{k - 1}} = \dfrac{3}{{k + 2}} = \dfrac{7}{{3k}}}$
$ { \Rightarrow \dfrac{2}{{k - 1}} = \dfrac{3}{{k + 2}}}$
$ { \Rightarrow 2k + 4 = 3k - 3}$
${ \Rightarrow k = 7}$
Hence, the value of k is $$7$$ .
Note: We knew the general equations , here we compare those with equations given in problem. After that applied the condition of infinite solution which is $$\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$$ .After comparing equation we got the value of k.
$ {{a_1}x + b{_1}y = {c_1}} $
$ {{a_2}x + b{_2}y = {c_2}} $
After comparing general equations with the given equation we got the value. After that we need to apply the condition of infinite solution which is $$\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$$ . after this we can get the value of k.
Complete step-by-step answer:
Consider the given equations.
2x + 3y = 7
(k - 1)x + (k + 2)y = 3k
The general equations
$ {{a_1}x + b{_1}y = {c_1}}$
$ {{a_2}x + b{_2}y = {c_2}}$
So,
$ {{a_1} = 2,{b_1} = 3,{c_1} = 7}$
${a_2}$ = k - 1,${b_2}$ = k + 2,${c_2}$ = 3k
We know that the condition of infinite solution
$$\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$$
Therefore,
${\dfrac{2}{{k - 1}} = \dfrac{3}{{k + 2}} = \dfrac{7}{{3k}}}$
$ { \Rightarrow \dfrac{2}{{k - 1}} = \dfrac{3}{{k + 2}}}$
$ { \Rightarrow 2k + 4 = 3k - 3}$
${ \Rightarrow k = 7}$
Hence, the value of k is $$7$$ .
Note: We knew the general equations , here we compare those with equations given in problem. After that applied the condition of infinite solution which is $$\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$$ .After comparing equation we got the value of k.
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