Difference between revisions of "005 Sample Final A, Question 9"

Revision as of 23:47, 18 May 2015

Question Solve the following system of equations

{\begin{aligned}2x+3y&=&1\\-x+y&=&-3\end{aligned}}

Step 1:
Add two times the second equation to the first equation. So we are adding $-2x+2y=-6$ to the first equation.
This leads to:
${\begin{array}{rcl}0+5y&=&-5\\-x+y&=&-3\end{array}}$

Step 2:
This gives us that $y=-1.$
Now we just need to find x. So we plug in -1 for y in the second equation.
${\begin{array}{rcl}-x-1&=&-4\\-x&=&-3\\x&=&3\end{array}}$

Final Answer:
$x=3,y=-1$

@@ Line 1: / Line 1: @@
 ''' Question ''' Solve the following system of equations <br>
-<center><math>  \begin{align} 2x + 3y  &= & 1\\ -x + y & = & -3\end{align}</math></center>
+::<math>  \begin{align} 2x + 3y  &= & 1\\ -x + y & = & -3\end{align}</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-! Final Answers
+! Step 1:
 |-
-|a) False. Nothing in the definition of a geometric sequence requires the common ratio to be always positive. For example, <math>a_n = (-a)^n</math>
+|Add two times the second equation to the first equation. So we are adding <math>-2x + 2y = -6</math> to the first equation.
 |-
-|b) False. Linear systems only have a solution if the lines intersect. So y = x and y = x + 1 will never intersect because they are parallel.
+|This leads to:
 |-
-|c) False. <math>y = x^2</math> does not have an inverse.
+|
+::<math>\begin{array}{rcl}
++ 5y &=& -5\\
+-x + y &=& -3
+\end{array}</math>
+|}
+{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+! Step 2:
 |-
-|d) True. <math>cos^2(x) - cos(x) = 0</math> has multiple solutions.
+|This gives us that <math>y = -1.</math>
 |-
-|e) True.
+|Now we just need to find x. So we plug in -1 for y in the second equation.
+|-
+|
+<math>\begin{array}{rcl}
+-x -1 &=& -4\\
+-x & =& -3\\
+x&=&3
+\end{array}</math>
+|}
+{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+! Final Answer:
 |-
-|f) False.
+|<math>x = 3, y = -1</math>
 |}