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Suppose I have these two algebraic equations in the format $y=mx+b$:

$$ y=2x+4 \\ y=3x+5 \\ $$

Now, by graphing these two algebraic equations on a coordinate plane, I find that they intersect at the point $(-1,2)$. Now, I find it annoying sometimes when trying to find where two lines intersect, I have two pull out a sheet of graph paper and plot them, or make a table. How can I find the intersection point of two algebraic equations without doing either of these things?

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    $\begingroup$ Solve $2x+4 = 3x+5$. That gets you the $x$ coordinate of the intersection (if any). Plug it in to get $y$. $\endgroup$ – Daniel Fischer Oct 22 '14 at 12:57
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At the point $(x_0, y_0)$ where they intersect, $(x_0, y_0)$ satisfies both equations, that is: $$ \left\{\begin{array}{rcl}y_0 = 2x_0 + 4 \\ y_0 = 3x_0 + 5 \end{array}\right. $$ In particular, both of the right-hand sides are equal to $y_0$, so they are equal to one another: $$2x_0 + 4 = 3x_0 +5.$$ Now, rearranging gives $$x_0 = -1.$$ Again, the point $(x_0, y_0)$ is on both lines, so we can substitute it in either equation, say, the first, and get a value for $y_0$: $$y_0 = 2(-1) + 4 = 2.$$ So, the intersection point is $$(x_0, y_0) = (-1, 2),$$ and we can check this, if we like, by substituting in the other equation.

It may be an instructive exercise, by the way, to work this out for a general pair of lines in the plane specified this way, that is, for the lines $$ \left\{\begin{array}{rcl}y &=& m\phantom{'} x + b\phantom{'} \\ y &=& m' x + b' \end{array}\right. $$ Note that when $m = m'$ there are either no solutions or infinitely many solutions---what do these special cases correspond to geometrically?

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The lines intersect when $2x+4 = 3x+5$. To find where your lines intersect, solve for $x$:

$$\begin{align} 2x+4 = 3x+5 &\iff -x = 1\\ &\iff x = -1 \end{align}$$

Now, solve for the y-coordinate by plugging $x = -1$ into either of the two equations:

Say, using $$y = 2x + 4 \implies y(-1) = 2$$

So the lines intersect at the point $(-1, 2)$.

Remark

Another approach is to simultaneously solve the system of two linear equations in two unknowns by subtracting the second equation from the first:

$$\begin{align} y & = 2x+4\\ (-)\quad y &= 3x+5\\ \hline 0 &= -x-1 \end{align}$$

Hence, $x = -1$. Then use either equation, as we did originally, to find the $y$-value corresponding to $x = -1$.

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Since the equations meet at some point so at that point the $y$ and $x$ values for both the straight line are similar hence

You just need to equate both of them:

$$2x+4=3x+5 \implies x=-1$$

This says that at $\,x=-1\,$ they both have same $x$ and $y$ value so using the second equation to solve for $y$ we get $y=3(-1)+5$

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