Answer:
6) x = 3, y = -1 or (3, -1)
8) x = -1, y = ½ or (-1, ½)
Step-by-step explanation:
Note:
I will only demonstrate problems 6 and 8 since the process of solving for the solution is practically similar for problems 6 and 7.
Given the following systems of linear equations with two variables:
Question 6
[tex]\displaystyle\mathsf{\left \{{{Equation\:1:\:x\:+\:y=2} \atop {Equation\:2:\:2x\:-\:y=7}} \right. }[/tex]
First, isolate y from Equation 1:
x + y = 2
x - x + y = -x + 2
y = -x + 2
Substitute the value of y in the previous step into Equation 2:
2x - (-x + 2) = 7
2x + x - 2 = 7
3x - 2 = 7
Add 2 to both sides:
3x - 2 + 2 = 7 + 2
3x = 9
Divide both sides by 3 to solve for x:
[tex]\displaystyle\mathsf{\frac{3x}{3}\:=\:\frac{9}{3}}[/tex]
x = 3
Substitute the value of x into Equation 1 to solve for y:
x + y = 2
3 + y = 2
3 - 3 + y = 2 - 3
y = -1
Therefore, the solution to the given system is: x = 3, y = -1 or (3, -1).
Question 8:
[tex]\displaystyle\mathsf{\left\{Equation\:1:\:3(6x\:-\:4y)\:=24} \atop{Equation\:2:\:3x\:-\:2y\:=4}} \right.}[/tex]
For Equation 2, divide both sides by -3:
[tex]\displaystyle\mathsf{\frac{-3(6x-4y)}{-3}=\frac{24}{-3} }[/tex]
6x - 4y = -8
Add 6x to both sides:
6x + 6x - 4y = 6x - 8
- 4y = 6x - 8
Divide both sides by -4 to isolate y:
[tex]\displaystyle\mathsf{\frac{-4y}{-4}=\frac{6x\:-\:8}{-4} }[/tex]
[tex]\displaystyle\mathsf{y=-\frac{3}{2}+2}[/tex]
Substitute the value of y from the previous step into Equation 2:
3x - 2y = -4
[tex]\displaystyle\mathsf{3x\:-\:2\Big(-\frac{3}{2}+2\Big)\:=-4}[/tex]
3x + 3 - 4 = -4
3x - 1 = -4
3x - 1 + 1 = -4 + 1
3x = -3
Divide both sides by 3 to solve for x:
[tex]\displaystyle\mathsf{\frac{3x}{3}=\frac{-3}{3}}[/tex]
x = -1
Substitute the value of x into Equation 2 to solve for y:
3x - 2y = -4
3(-1) - 2y = -4
-3 - 2y = -4
Add 3 to both sides:
-3 + 3 - 2y = -4 + 3
-2y = -1
Divide both sides by -2 to isolate y:
[tex]\displaystyle\mathsf{\frac{-2y}{-2}=\frac{-1}{-2}}[/tex]
y = ½
Therefore, the solution to the given system is: x = -1, y = ½ or (-1, ½).
