Question

Solve the system of linear equations below.  2x + 3y = 2   x + 6y = 4

Answer 1

2x+3y=2
x+6y=4
Multiply equation II by II
2x+3y=2
2x+12y=8
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9y=6
y=6/9
y=2/3
Solving for x by replacing in equation 1;
2x+3*2/3=2
2x+2=2
2x=0
x=0

Answer 2

Answer:

$(0, \frac{2}{3})$

Step-by-step explanation:

Given the system of equation:

2x + 3y = 2            ....[1]

x + 6y = 4          .   ...[2]

Multiply by 2 to both sides in [1] we get;

4x + 6y =4           ....[3]

Subtract equation [2] from [3] to eliminate y we get;

$4x-x = 0$

⇒$3x = 0$

Divide both sides by 3 we have;

x = 0

Substitute the value of x in [1] we have;

2(0)+ 3y = 2

⇒$0+3y =2$

⇒$3y = 2$

Divide both sides by 3 we have;

$y = \frac{2}{3}$

Therefore, the solution for the given system of equation is, $(0, \frac{2}{3})$