Answer:
Q1: No solutions
Q2: x=2, y=-2
Q3: x=1, y=4
Step-by-step explanation:
Q1: Found out 0y = 18 which is impossible therefore no solution.
Q2:
In the first equation: [tex]2x - 4y = 12[/tex]
Let's express x in terms of y by changing the subject. Follow my steps.
[tex]2x = 12+4y[/tex]
[tex]x = \frac{12+4y}{2}[/tex]
Now let's put the x above into the second equation: [tex]-14x-4y=-20[/tex]
It becomes: [tex]-14(\frac{12+4y}{2})-4y=-20[/tex]. Follow my steps.
[tex]-7(12+4y)=-20+4y[/tex]
[tex]-84-28y=-20+4y[/tex]
[tex]-32y=64[/tex]
[tex]y=-2[/tex]
Put y = -2 into the first equation:
[tex]2x-4(-2)=12[/tex]
[tex]2x+8=12[/tex]
[tex]2x=12-8[/tex]
[tex]2x=4[/tex]
[tex]x=2[/tex]
Q3:
Multiply the first equation by 3 and the second equation by 7, so that both are 21x.
1st equation:
[tex]3(7x-3y)=3(-5)[/tex]
[tex]21x-9y=-15[/tex]
2nd equation:
[tex]7(3x+2y) = 7(11)[/tex]
[tex]21x+14y = 77[/tex]
We can now substract the 2nd equation with the 1st equation so that both 21x cancels out and only the variable y remains.
[tex]21x-9y-(21x+14y)=-15-77[/tex]
[tex]21x-9y-21x-14y=-92[/tex]
[tex]-23y=-92[/tex]
[tex]y=4[/tex]
Put y = 4 into the second equation to find x.
[tex]3x+2(4)=11[/tex]
[tex]3x+8=11[/tex]
[tex]3x=11-8[/tex]
[tex]3x = 3[/tex]
[tex]x=1[/tex]
