Step by step
The question
Question: for which k do a + 2b = 3 and 4a + kb = 1 have no solution?
We call them ① and ②.
No solution means no point (x, y) works for both.
1 Make the a parts the same
To compare the equations, we make the parts of one letter the same.
k is with b, so we use the other letter, a: it has plain numbers in front.
The a parts: a in ① and 4a in ②.
4 divided by 1 is 4.
So a times 4 is 4a.
We multiply every part of ① by 4.
We do it on both sides, so they stay equal.
4 times a is 4a.
4 times 2b is 8b.
4 times 3 is 12.
We call it ③.
2 Compare ③ and ②
We write ② again, under ③.
If the left sides are the same, both equations talk about the same amount, 4a + 8b.
One amount cannot be 12 and 1 at once.
So no point works: no solution.
If the numbers are the same too, it is one equation written twice.
Every point on that line works: infinitely many solutions.
If the left sides are different, we can take one equation away from the other.
Some b is left, so we can find b.
That gives one solution.
Careful!
We compare with ③, not ①.
① was times 4.
2b times 4 is 8b.
3 Find k
We want no solution.
So the left sides must be the same: kb = 8b.
k times b is 8 times b when k is 8.
So k = 8.
And the numbers, 12 and 1, are different.
Good.
4 Check
Let's check.
We put k = 8 into ②.
Take ② away from ③.
The left sides are the same, so together they make 0.
The numbers: 12 take away 1.
0 = 11 is never true.
So there is no solution.
It works!
On a grid, the two lines look like this.
Same x and y parts means the same slope.
So the lines are parallel.
Parallel lines never meet: no solution.
Let's look back.
First: make the a parts the same.
Then: compare ③ and ②.
Then: find k.
Last: check.