When given a graph, draw another line to solve an equation
Using a graph is just another way we can solve equations. We can solve linear, quadratic and even cubic ( see http://cribbd.com/learn/maths/algebra/solvecubicequationsbydrawingappropriatelinesongraphs for cubic graphs).
Linear Equations Example: Say we wanted to solve 2x1 = 3x+1, and we are given the equation y = 2x1 on a graph.
In order to solve the equation for 2x1 = 3x+1, we need to plot y = 3x+1 on the same graph, and see where both lines intercept one another. We then take the x coordinate as the solution for the equation.
So looking at the graph we can see that the lines intersect one another at x=2
Quadratic Equation Example: Say we have the graph x^{2 }4x12 given to us, and we need to solve x^{2 }4x12 = x+4.
To do this, we need to plot y = x+4 on the same set of axis, then read off where y=x+4 intersects the quadratic.
We can give estimates to where the intersecting points lie on the xaxis. So one at x= 2.1 and the other at x = 7.2. Remember since this is a quadratic graph we are solving, there are two possible values can be.
To check this, we need only to put the values back into our equation x^{2 }4x12 = x+4.
If x = 2.2, on one side of the equation we have (2.2)^{2}  (4 x  2.2)  12 = 1.64. On the other hand side we have 2.2 + 4 = 1.8. 1.64 and 1.8 are quite close together so we can say that x=2.2 is a solution.
If x=7.2, on one side of the equation we have 7.2^{2} 4 x 7.2  12 = 11.04. On the other hand side we have 7.2 + 4 = 11.2. Again these two values are quite close together, so we can say that x=7.2 is a solution
As you can see, unless a line goes through a graph where x is a whole number, an estimate has to be made. You need to check that both sides are fairly similar when going to check your answers.
Nothing in this section yet. Why not help us get started?
Related Topics
Requires a knowledge of…
Related Questions

1Vote3Answers

1Vote2Answers

1Vote3Answers

0Votes4Answers

1Vote2Answers

0Votes1Answer

0Votes1Answer

2Votes4Answers

0Votes2Answers