a + 2b = 2 ................(1)

2a -3b = 4 .................(2)

We will use the elimination method to solve.

We will multiply (1) by -2 and add to (2).

==> -2a -4b = -4 .....................-2*(1)

==> 2a -3b = 4 .....................(2)

==> -7b = 0 ==> b= 0

Now to calculate a we will substitute into (1).

==> a + 2b = 2 ==> a + 0 + 2 ==> a = 2

==> a = 2

Then the answer to the system is :

**a = 2 and b= 0**

We have to solve for a and b given a + 2b = 2 and 2a -3b = 4

a + 2b = 2

=> a = 2 - 2b

Substitute this in 2a - 3b = 4

=> 2( 2 - 2b) - 3b = 4

=> 4 - 4b - 3b = 4

=> -7b = 0

=> b = 0

a = 2 - 2b = 2

**Therefore a = 2 and b = 0**

Q;

a+2b = 2.......(1)

2a -3b = 4 ....(2)

We have to find a and b.

Solution:

2*(1)-(2): 2(a+2b) - (2a-3b) = 2*2-4 = 0.

4b+3b = 0.

7b = 0.

7b/7 = 0/7.

b= 0.

We put b= 0 in (1): a+2b = 2. So a+2*0 = 2. So a = 2.

a = 2 and b = 0.

**a = 2. b = 0.**