Question

How to find \[{x_1},{x_2},{x_3}\]
\[5{x_1} + 3{x_2} + 2{x_3} = 4,\] \[3{x_1} + 3{x_2} + 2{x_3} = 2,\] \[{x_2} + {x_3} = 5.\]

Answer 1

Hint: To solve linear equations, we will multiply these equations with some constants and then equate equations to find the values of \[x's\] .

Formula Used:
If two equations are given in the below format:
\[Ax + By = C\]
\[Mx + Ny = D\]
Then we can equate both the equations using the value of \[x\] .
To get the value of \[x\] from the first equation, we can perform the following steps:
\[Ax + By = C\]
\[ \Rightarrow Ax = C - By\]
\[ \Rightarrow x = \dfrac{{C - By}}{A}\]
Again, we can do the same operation for the second equation also:
\[Mx + Ny = D\]
\[ \Rightarrow Mx = D - Ny\]
\[ \Rightarrow x = \dfrac{{D - Ny}}{M}\]
Now, if we equate these two above equations, we can find the value of \[y\] from the above equations.
Like,
\[\dfrac{{D - Ny}}{M} = \dfrac{{C - By}}{A}\]
\[ \Rightarrow AD - ANy = CM - BMy\]
\[ \Rightarrow y(BM - AN) = (CM - AD)\]
\[ \Rightarrow y = \dfrac{{(CM - AD)}}{{(BM - AN)}}.\]
Now, putting these values of \[y\] , in any of the above equations, we can find the value of \[x\] .

Complete step by step answer:
In the above question following three equations are given:
\[5{x_1} + 3{x_2} + 2{x_3} = 4.................(1)\]
\[3{x_1} + 3{x_2} + 2{x_3} = 2.................(2)\]
And, \[{x_2} + {x_3} = 5.................(3)\]
Now, if we observe the first two equations, it is clearly visible that \[3{x_2} + 2{x_3}\] term is common in both the equations.
So, if we subtract the second equation from the first equation, we can find the value of \[{x_1}\] .
So, by doing \[\{ eq(1) - eq(2)\} \] , we get:
\[ \Rightarrow (5{x_1} + 3{x_2} + 2{x_3}) - (3{x_1} + 3{x_2} + 2{x_3}) = (4 - 2)\]
On simplify we get,
\[ \Rightarrow 2{x_1} = 2\]
By dividing both sides by \[2\] , we get:
\[ \Rightarrow {x_1} = \dfrac{2}{2} = 1.\]
Now, if we put this value of \[{x_1}\] in the second equation, we get:
\[ \Rightarrow (3 \times 1 + 3{x_2} + 2{x_3}) = 2\] .
Now, by performing the calculations, we get:
\[ \Rightarrow (3{x_2} + 2{x_3}) = 2 - 3 = - 1.............(4)\]
Now, we will multiply third equation by \[3\] , we get:
\[ \Rightarrow (3{x_2} + 3{x_3}) = 15..........(5)\]
Now, if we subtract equation \[(4)\] from equation \[(5)\] , we get:
\[ \Rightarrow (3{x_2} + 3{x_3}) - (3{x_2} + 2{x_3}) = 15 - ( - 1)\] .
Now, by doing further simplification:
\[ \Rightarrow {x_3} = 16\] .
Now, putting the value of \[{x_1} = 1\] and \[{x_3} = 16\] in equation \[(1)\] , we get:
\[ \Rightarrow 5 \times 1 + 3{x_2} + 2 \times 16 = 4\] .
Now, by doing the multiplication, we get:
\[ \Rightarrow 5 + 3{x_2} + 32 = 4\]
Let us add the term and we get,
\[ \Rightarrow 3{x_2} + 37 = 4\] .
Now, taking the constant term into the R.H.S and performing further calculations, we get:
\[ \Rightarrow 3{x_2} = 4 - 37\]
On subtracting we get
\[ \Rightarrow 3{x_2} = - 33\]
On dividing we get
\[ \Rightarrow {x_2} = - \dfrac{{33}}{3} = - 11.\]

Therefore, the solutions of the given equations are: \[{x_1} = 1,{x_2} = - 11,{x_3} = 16.\]

Note: If the coefficients of variables of the equations are equal in any linear equation, then it is not possible to find the values of the variables, as it shows the infinite solution.