To state: The solutions for the system of equation by elimination also check the answer.
{x+2y=22x+3y−z=−94x+2y+5z=1
The resultant solution is (−12211,7211,7111) .
Given information:
The given system equations is:
{x+2y=22x+3y−z=−94x+2y+5z=1
Explanation:
Write the original system equations:
{x+2y=22x+3y−z=−94x+2y+5z=1
Let’s number the equation,
{x+2y=2⋯(1)2x+3y−z=−9⋯(2)4x+2y+5z=1⋯(3)
Multiply equation (2) by -5,
(2x+3y−z=−9)×510x+15y−5z=−45⋯(4)
Add equations (3) and (4) to eliminate variable z ,
{4x+2y+5z=110x+15y−5z=−45¯14x+17y=−44
Number the resultant equation,
14x+17y=−44⋯(4)
Now multiply equation (1) by -14,
(x+2y=2)×−14−14x−28y=−28⋯(5)
Add equation (4) and (5) to eliminate variable x ,
{−14x−28y=−2814x+17y=−44¯ −11y=−72y=7211
Now find the value of variable x by substituting the value of y in equation (1) and simplify,
x+2y=2x+2(7211)=2x=2−14411x=22−14411x=−12211
Now find the value of variable z by substituting the values of x and y in equation (2) and simplify,
2x+3y−z=−92(−12211)+3(7211)−z=−9−24411+21611−z=−9−2811−z=−9−z=−9+2811−z=−7111z=7111
Therefore, the solution for the given system of equation is (−12211,7211,7111) .
Now check the solution by substituting the values of variables x , y and z into the other two original equations,
x+2y=2(−12211)+2(7211)?=2−122+14411?=22211?=22=2
Similarly substitute values of variables x , y and z into equation (3),
4x+2y+5z=14(−12211)+2(7211)+5(7111)?=1−48811+14411+35511?=1−488+144+35511?=11111?=11=1
Chapter 3 Solutions
High School Math 2015 Common Core Algebra 2 Student Edition Grades 10/11
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