[amp_mcq option1=”unique solution” option2=”infinite number of solutions” option3=”no solution” option4=”exactly two solutions” correct=”option1″]<\/p>\n
The correct answer is: A. unique solution<\/p>\n
To solve a system of equations, we can use the elimination method. In this method, we eliminate one variable at a time by adding or subtracting the equations in a way that cancels out the variable.<\/p>\n
For the system of equations $x + 2y + z = 6$, $2x + y + 2z = 6$, and $x + y + z = 5$, we can eliminate $z$ by adding the first two equations together. This gives us $3x + 3y = 12$. We can then eliminate $y$ by subtracting the third equation from the second equation. This gives us $x = 1$. Substituting $x = 1$ into the first equation, we get $1 + 2y + z = 6$. Solving for $y$, we get $y = 2$. Substituting $x = 1$ and $y = 2$ into the third equation, we get $1 + 2 + z = 5$. Solving for $z$, we get $z = 2$.<\/p>\n
Therefore, the system of equations has a unique solution $(x, y, z) = (1, 2, 2)$.<\/p>\n
Here is a brief explanation of each option:<\/p>\n
[amp_mcq option1=”unique solution” option2=”infinite number of solutions” option3=”no solution” option4=”exactly two solutions” correct=”option1″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[489],"tags":[],"class_list":["post-19996","post","type-post","status-publish","format-standard","hentry","category-linear-algebra","no-featured-image-padding"],"yoast_head":"\n