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Solve system -3-9-1-3-x-0 give solution real form -x2-q 17074295

Question:

Solve system {eq}-3-9-1-3-x-0{/eq} give solution real form {eq}-x2-q{/eq} {eq}17074295{/eq}

Elimination method:

Using the following steps, we can solve the system of equations using elimination:

Step 1:

Multiply each equation by some suitable non-zero constant to make the coefficients of one variable equal to each other.

Step 2:

Add or subtract the two resulting equations so that one variable gets eliminated. If you get an equation in one variable, go to Step 3.

If in Step 2, we obtain a true statement involving no variable, then the original pair of equations has infinitely many solutions.

If in Step 2, we obtain a false statement involving no variable, then the original pair of equations has no solution, i.e., it is inconsistent.

Step 3:

Solve the equation in one variable to find the value of that variable.

Step 4:

Substitute this value into either of the original equations to get the value of the other variable.

Answer and Explanation:

Example;

{eq}\displaystyle \eqalign{ & (a)\,\,{x_1} + 2{x_2} + 3{x_3} = - 1\,,\,4{x_1} + 5{x_2} + 3{x_3} = 2 \cr & {\text{Let, }}{x_3} = k,\,\,(k\,{\text{is real number)}} \cr & {x_1} + 2{x_2} = - 1\, - 3k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1) \cr & 4{x_1} + 5{x_2} = 2 - 3k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (2) \cr & 4{\text{Equation(1) - Equation(2);}} \cr & 4({x_1} + 2{x_2}) - (4{x_1} + 5{x_2}) = 4( - 1\, - 3k) - (2 - 3k) \cr & 4{x_1} + 8{x_2} - 4{x_1} - 5{x_2} = - 4\, - 12k\, - 2 + 3k \cr & 4{x_1} - 4{x_1} + 8{x_2} - 5{x_2} = - 4\,\, - 2 - 12k + 3k \cr & 4{x_1} - 4{x_1} + 8{x_2} - 5{x_2} = - 6\, - 9k \cr & 3{x_2} = - 6 - 9k\,\, \Rightarrow \,3{x_2} = - 3(2 + 3k) \cr & \,3{x_2} = - 3(2 + 3k)\, \Rightarrow \,{x_2} = - (2 + 3k) \cr & {\text{From equation(1);}} \cr & {x_1} + 2{x_2} = - 1\, - 3k \cr & {x_2} = - (2 + 3k) \cr & {x_1} - 2(2 + 3k) = - 1\, - 3k \cr & {x_1} - 4 - 6k = - 1\, - 3k \cr & {x_1} = - 1\, - 3k + 6k + 4 \cr & {x_1} = 3k + 3 \cr & {x_1} = 3(k + 1) \cr & {\text{Solution of the equation;}} \cr & {x_1} = 3(k + 1)\,\,,{\text{ }}{x_2} = - (2 + 3k)\,\,{\text{and }}{x_3} = k,\,k \in R \cr & \text{There exist an infinite number of solutions} & \cr & \cr & (b)\,\,2{x_1} - 4{x_2} + 2{x_3} = - 6\,,\,{x_1} + 2{x_2} - {x_3} = 3 \cr & {\text{Let, }}{x_3} = k,\,\,(k\,{\text{is real number)}} \cr & 2{x_1} - 4{x_2} = - 6\, - 2k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1) \cr & {x_1} + 2{x_2} = 3 + k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (2) \cr & {\text{Equation(1) - 2Equation(2);}} \cr & 2{x_1} - 4{x_2} - 2({x_1} + 2{x_2})\,\, = - 6\, - 2k\, - 2(3 + k)\,\, \cr & 2{x_1} - 4{x_2} - 2{x_1} - 4{x_2} = - 6\, - 2k\,\, - 6 - 2k \cr & 2{x_1} - 2{x_1} - 4{x_2} - 4{x_2} = - 6\, - 6 - 2k - 2k \cr & - 8{x_2} = - 12 - 4k\,\, \Rightarrow \, - 8{x_2} = - 4(3 + k) \cr & - 8{x_2} = - 4(3 + k)\,\, \Rightarrow \,{x_2} = \frac{1}{2}(3 + k) \cr & {\text{From eqution(1);}} \cr & 2{x_1} - 4{x_2} = - 6\, - 2k\,\,\, \cr & {x_2} = \frac{1}{2}(3 + k) \cr & 2{x_1} - 4\frac{1}{2}(3 + k) = - 6\, - 2k \cr & 2{x_1} - 2(3 + k) = - 6\, - 2k\, \cr & 2{x_1} - 6\, - 2k\, = - 6\, - 2k\,\, \cr & 2{x_1} = 0 \cr & {x_1} = 0 \cr & {\text{Solution of the equation;}} \cr & {x_1} = 0\,\,,{\text{ }}{x_2} = \frac{1}{2}(3 + k)\,\,\,{\text{and }}{x_3} = k\,,\,k \in R \cr & {\text{For both of the problems there exist infinitely many solutions}}{\text{.}} \cr} {/eq}


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Elimination Method in Algebra: Definition & Examples

from High School Algebra II: Help and Review

Chapter 7 / Lesson 9
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