Question

How do you solve \[y = x + 4\] and \[y = - 2x + 1\] by graphing?

Answer 1

Hint: We have a system of two linear equations and we need to solve it by graphing. First we draw the graph for \[y = x + 4y\] and then for \[y = - 2x + 1\] in the same coordinate system. The two lines intersect at one point and it will be the solution for the given equations.

Complete step by step solution:
Given, \[y = x + 4\] and \[y = - 2x + 1\] .
Now we take
 \[y = x + 4\]
We use intercept form to graph the equation.
Put \[x = 0\] in \[y = x + 4\] , we will have y intercept.
 \[y = 0 + 4\]
 \[ \Rightarrow y = 4\] .
Thus we have one coordinate point \[\left( {0,4} \right)\] .
Put \[y = 0\] in \[y = x + 4\] , we will have x intercept.
 \[0 = x + 4\]
 \[ \Rightarrow x = - 4\] .
Thus we have another coordinate point \[\left( { - 4,0} \right)\] .
Thus for \[y = x + 4\] we have two coordinate points \[\left( {0,4} \right)\] and \[\left( { - 4,0} \right)\] .
Similarly, we take \[y = - 2x + 1\] .
We use intercept form to graph the equation.
Put \[x = 0\] in \[y = - 2x + 1\] , we will have y intercept.
 \[y = - 2(0) + 1\]
 \[ \Rightarrow y = 1\] .
Thus we have one coordinate point \[\left( {0,1} \right)\] .
Put \[y = 0\] in \[y = - 2x + 1\] , we will have x intercept.
 \[0 = - 2x + 1\]
 \[2x = 1\]
 \[ \Rightarrow x = \dfrac{1}{2}\] .
 \[ \Rightarrow x = 0.5\]
Thus we have another coordinate point \[\left( {0.5,0} \right)\] .
Thus for \[y = x + 4\] we have two coordinate points \[\left( {0,1} \right)\] and \[\left( {0.5,0} \right)\] .
Now plotting the graph we have,

As we can see in the graph that the two lines intersect at \[\left( { - 1,3} \right)\] . Hence the solution of \[y = x + 4\] and \[y = - 2x + 1\] is \[x = - 1\] and \[y = 3\] .

Note: We can solve this using elimination method or by substitution method or by cross multiplication method. In any case we will have the same answer. If we have a quadratic equation and if we draw the graph of the quadratic solution then the solution is ‘x’ intercept and ‘y’ intercept.