Question

Y= 6X +15 and Y =MX - 4.
X= - 2
Find the Y intercept point must work for both

Answer 1

At
`\(X = -2\)` , when
`\(Y = 3\)` from
`\(Y = 6X + 15\)` , equating with
`\(Y = MX - 4\)`,
`\(M = -(7)/(2)\)`. The y-intercept point that satisfies both equations is
`\((-2, 3)\).`

Given equations:

1.
`\(Y = 6X + 15\)`

2.
`\(Y = MX - 4\)`

And the value
`\(X = -2\)`.

Substitute
`\(X = -2\)` into equation 1 to find the corresponding
`\(Y\)`:

`\[Y = 6X + 15\]`

`\[Y = 6(-2) + 15\]`

`\[Y = -12 + 15\]`

`\[Y = 3\]`

So, according to equation 1, when
`\(X = -2\),`
`\(Y = 3\).`

Substitute
`\(X = -2\)` into equation 2 and set it equal to the previously found value of
`\(Y\)`:

`\[Y = MX - 4\]`

`\[3 = M(-2) - 4\]`

Now, solve for
`\(M\)`:

`\[3 = -2M - 4\]`

Add 4 to both sides:

3 + 4 = -2M

7 = -2M

Divide both sides by -2:

`\[M = (7)/(-2)\]`

`\[M = -(7)/(2)\]`

Therefore, for
`\(X = -2\)` to satisfy both equations, the value of
`\(M\)`should be
`\(-(7)/(2)\)`, and the y-intercept point is
`\((-2, 3)\)`.