x+y+z=6
x-y-z=4
I'm going to *guess* that your issue is about how to describe a line in $\displaystyle \mathbb{R}^3$. There are two equivalent ways: parametric equations and vectors.
When planes intersect, they intersect in a line, right? And how is a line described in $\displaystyle \mathbb{R}^3$? I like to think of it in terms of vectors.
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Let $\displaystyle \vec{p}$ be a vector from the origin to any point on the line. Let $\displaystyle \vec{q}$ be a vector parallel to the line.
Then any point on the line is gotten to by moving from the origin to $\displaystyle \vec{p}$, and then moving (i.e. vector addition)
some multiple t of $\displaystyle \vec{q}$. (Note that t negative means moving in the opposite direction of $\displaystyle \vec{q}$.)
Thus $\displaystyle Line = \{ \vec{p} + t \vec{q} \in \mathbb{R}^3 \ | \ t \in \mathbb{R} \}$.
Notice that there's 1 free parameter there, the t. One number, t, tells you where you are the line.
In terms of coordinates, if $\displaystyle \vec{p} = (p_1, p_2, p_3)$ and $\displaystyle \vec{q} = (q_1, q_2, q_3)$, then
$\displaystyle Line = \{ (p_1 + q_1t, p_2 + q_2t, p_3 + q_3t) \in \mathbb{R}^3 \ | \ t \in \mathbb{R} \}$.
Notice how, in each coordinate, you get something that looks like the "mx+b" equation of the line in the plane?
That observation leads to the parametric form for writing a line in 3-space:
$\displaystyle Line = \{ (x, y, z) \in \mathbb{R}^3 \ | x = p_1 + q_1t, y = p_2 + q_2t$, and $\displaystyle z = p_3 + q_3t$, all for the same $\displaystyle t \in \mathbb{R} \}$
or just "the line is given by parametric equations": $\displaystyle x(t) = q_1t + p_1, y(t) = q_2t + p_2$, and $\displaystyle z(t) = q_3t + p_3$ where $\displaystyle t \in \mathbb{R}$.
Notice again the one free parameter t.
So if your goal is to find a line, then you're hoping to find x, y, and z (whether vector or parametrically) all in terms of one free parameter.
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One final observation. Sometimes, the free parameter is x itself (or y, or z, but USUALLY x - HOWEVER, in this problem it actually won't be x).
That's because you could take your solution, and use that $\displaystyle x(t) = p_1 + q_1t$ to find t in terms of x (assuming $\displaystyle q_1 \ne 0$ - BEWARE, in this problem it will be, but usually it won't).
So if $\displaystyle q_1 \ne 0$, then $\displaystyle t = (x-p_1)/q_1$, so $\displaystyle y = p_2+ q_2((x-p_1)/q_1)$ and $\displaystyle z = p_3+ q_3((x-p_1)/q_1)$, and, after simplifying, you'll get $\displaystyle y=m_1x+b_1, z = m_2x + b_2$.
Then the line would be $\displaystyle Line = \{ (x, m_1x+b_1, m_2x + b_2) \in \mathbb{R}^3 \ | \ x \in \mathbb{R} \}$.
The thing to notice is that, even in this case, there's still exactly 1 free parameter - only now it's x, not t.
The 1 free parameter is what tells you you've discovered a line in 3-space!
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So with that in mind, consider a point (x,y,z) in the intersection of those planes. Being in both planes, x, y, and z satisfy both those equations for a plane. Those are 2 equations in 3 unknowns. After doing some algebra (solve in one equation, plug into the other), you'll have 1 equation in 2 unknowns. But one equation is 2 unknowns is just like for a line (y = mx+b). You treat one of those as a free parameter, and then you've an equation for one of your coordinates in term of that free parameter. Plugging back into one of the original equations tells you the other coordinate in terms of that free parameter. Then you'll have all your coordinates in terms of one free parameter, hence you'll have described a line, and so found the intersection of those two planes.
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A final observation. Consider the equation for a plane given by x-y-z=4. Why should the points that satisfy that make up a plane? Because it's a way of saying that any point in the plabe can be located by providing TWO free parameters: x and y. Just to clarify what's a coordinate, and what;s a free parmeter, I'll use s and t instead of x and y.
Then given values for the TWO free parameters s and t, you get a point on the plane (x, y, z), according to:
x(s,t) = s, y(s, t) = t, z(s, t) = x(s,t) - y(s,t) - 4 = s - t - 4.
Parametric for that plane: $\displaystyle x(s,t) = s, y(s, t) = t, z(s, t) = s - t - 4$, where s, t \in $\displaystyle \mathbb{R} \}$.
Coordinates for that plane: $\displaystyle Plane = \{ ( s, t, s - t - 4) \in \mathbb{R}^3 \ | \ s, t \in \mathbb{R} \}$.
To go to vector form, note ( s, t, s - t - 4) = s(1, 0, 1) + t(0, 1, -1) + (0, 0, -4),
Let \vec{v_1} = (1, 0, 1), \vec{v_2} = (0, 1, -1), \vec{v_1} = (0, 0, -4). Then
Vector form for that plane: $\displaystyle Plane = \{ s\vec{v_1} + t\vec{v_2} +\vec{v_3} \in \mathbb{R}^3 \ | \ s, t \in \mathbb{R} \}$.
Notice how it's 2 free parameters, s and t, in each case? "1 free parameter = line, 2 free parameters = plane". That's not a coincidence!!
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When we identify our two free parameters with x and y (like how for the line we used x itself as a free paramter instead of t)(and again, not always possible - IT WON'T BE POSSIBLE in this particular problem), then the 3 methods above become:
Parametric for that plane: $\displaystyle z(x, y) = x - y - 4$, where x, y \in $\displaystyle \mathbb{R} \}$. (Note no need to write x = x, y=y).
Coordinates for that plane: $\displaystyle Plane = \{ ( x, y, x - y - 4) \in \mathbb{R}^3 \ | \ x, y \in \mathbb{R} \}$. (Note it's "above" (x,y) in the plane.)
Vector form for that plane doesn't change meaningfully: $\displaystyle Plane = \{ x\vec{v_1} + y\vec{v_2} +\vec{v_3} \in \mathbb{R}^3 \ | \ x, y \in \mathbb{R} \}$.
Notice how the parametric for the plane became one equation, Parametric for that plane: $\displaystyle z = x - y - 4$ - exactly what we started with.
Eliminate z between the two equations, you get:
x=5
Substitute back in one of the equations and find:
y+z=1
So, intersection line is {x=5, y+z=1}