Question

Could someone help me, please

Answer 1

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: $2x - 4y = 12$

Let's express x in terms of y by changing the subject. Follow my steps.

$2x = 12+4y$

$x = \frac{12+4y}{2}$

Now let's put the x above into the second equation: $-14x-4y=-20$

It becomes: $-14(\frac{12+4y}{2})-4y=-20$. Follow my steps.

$-7(12+4y)=-20+4y$

$-84-28y=-20+4y$

$-32y=64$

$y=-2$

Put y = -2 into the first equation:

$2x-4(-2)=12$

$2x+8=12$

$2x=12-8$

$2x=4$

$x=2$

Q3:

Multiply the first equation by 3 and the second equation by 7, so that both are 21x.

1st equation:
$3(7x-3y)=3(-5)$
$21x-9y=-15$

2nd equation:

$7(3x+2y) = 7(11)$

$21x+14y = 77$

We can now substract the 2nd equation with the 1st equation so that both 21x cancels out and only the variable y remains.

$21x-9y-(21x+14y)=-15-77$

$21x-9y-21x-14y=-92$

$-23y=-92$

$y=4$

Put y = 4 into the second equation to find x.

$3x+2(4)=11$

$3x+8=11$

$3x=11-8$

$3x = 3$

$x=1$