Q&A for How to Solve Simultaneous Equations Graphically

If the graphed lines do not intersect, the system has no solution. That means there is no set of values for the variables that will work for both equations simultaneously.

You don't guess any numbers. You simply use a low number for x (like 0, 1, 2, -1, or -2, for example) so that it's easy to calculate the corresponding value of y.

Plot both equations on the same graph. The x and y values of the point where the two lines intersect are the solution of the equations.

Choosing zero for the value of x simply makes it easier and quicker to find the corresponding value of y, which would be the y-intersect.

First algebraically solve for y in terms of x (or vice versa). Then assign convenient values to x, and calculate corresponding values of y. Then plot some pairs of values (at least two pairs for a straight-line graph; more than two for a curved-line graph), and draw the graph. (You can graphically find the value of one variable when zero is assigned as the value of the other variable by simply noticing where the graph crosses each of the axes.)

All non-parallel lines (in the same plane) will meet somewhere. Two lines that are parallel to each other will not meet. That means that they have the same slope. In that case the two equations are not solvable.

Yes. Any "linear" equations like these (meaning relatively simple equations) are solvable this way. The one exception occurs when the equations graph as parallel lines, in which case the lines don't intersect.

Yes. It doesn't matter what values are used in plotting the lines. Once plotted, they will intersect in one point only (assuming straight lines) regardless of the values used, and the point of intersection is the solution to the problem.

By plotting each of those equations, you will get two straight lines which intersect at the point (3,4). That means that both equations are true when x=3 and y=4. Graphically, the solution to this system of equations is (3,4).

18 and -8 are the x- and y-coordinates, respectively, of the point of intersection of the two lines in the example graph.

To solve that system graphically: First solve each equation for x in terms of y, or y in terms of x (it doesn't matter which). Then make two tables of values (one for each equation) by assigning convenient values to the independent variable and calculating the corresponding values for the dependent variable. (For example, solve the second equation x: x = 3y + 1. Here y is the independent variable, and x is the dependent variable. Assign small values to y, and find the corresponding values for x. Then plot a line for each equation using at least two pairs of values for each. Both will be straight lines. The system's solution consists of the coordinates of the lines' point of intersection.

First re-write each equation to set y equal to an expression in x (or vice versa). (In this case, y = -1 - x, and y = 5 - 2x.) Then you can solve the system graphically (as shown above) or algebraically (by substitution or elimination).

If x = 1/3, plot it as a vertical line crossing the x-axis at +1/3. If y = 1/3, plot it as a horizontal line crossing the y-axis at +1/3.

Yes, you can (assuming the equations are already linear), but substituting zero is a really easy method.

It doesn't matter which values you choose for x (or the independent variable). If you perform the calculations correctly, you'll get the appropriate corresponding values of y (or the dependent variable), and the graph will be correct.

Graph each equation on the same grid. The point or points of intersection of the two graphed lines will be the solution(s) expressed as values of x and y. (If the lines don't intersect, there is no solution.)

This is a relatively simple exercise. Solve for y in terms of x in both equations. Then plot the two straight lines corresponding to the equations. Look closely at the point of intersection of the two lines. The x- and y-coordinates of that point are the x- and y-values you're looking for.

Those two equations would be represented by parallel lines on a graph. So the lines would never intersect, and there is no solution to that system. In other words, there are no values of x and y that would satisfy both equations simultaneously.

First combine the like terms. x + 32 = 14. Then subtract 32 on both sides, leaving you with + x = 14 - 32. 14 - 32 = - 18. So x = -18.

Either follow the instructions above, or use the elimination method, which is relatively simple in this case: First subtract either equation from the other in order to get rid of the y terms. For example, subtract the first equation from the second to find that x = 2. Plug that x-value into either original equation to find that y = 3.

It's easiest to use the elimination method: Add these two equations together. You get 2x = 6, and x = 3. Plug that x-value into either original equation, and you find that y = 1.

Use the elimination method (often the easiest method): first multiply the first equation by 2 to get 6x - 2y = 8. Then add that equation to the second original equation in order to eliminate the y-terms: 7x = 14, and x = 2. Plug that x-value into either original equation to solve for y: 3(2) - y = 4, so 6 - y = 4, and y = 2. Both x and y equal 2.

First set up a graph with r and s axes (instead of x and y). Then solve both equations for r in terms of s. (If you prefer, you could solve for s in terms of r.) Now create a table of values with corresponding values of r and s (as you would with x and y). Then draw a graph of the two lines representing the equations. The r- and s-coordinates of the point of intersection of the two lines are the r- and s-values you're looking for.

Since the variables are the same, you can set these two equations on opposite sides of an equals sign (2x-5 = x+4). Subtract x on both sides and add 5 to both sides (x=9). Use that value of x and input it into the simpler y=x+4 equation so you get a y value of y=9+4. You'll get y=13.

Assuming you want to use the graphical method, you would simply graph both equations. Look at the point where the two lines intersect. The x- and y-coordinates of the point of intersection are the x- and y-values you're looking for. (Either the substitution method or the elimination method would probably solve the equations more easily in this particular case.)

Because there are two unknowns, you need two equations to solve this problem. One equation is not enough. All you can do is solve for a variable in terms of another variable: y = 3x + 2 or x = 2/-3 - y/3.

Solve by the elimination method. (You make that choice by noticing that one of the equations has the term +2x², while the other equation has the term -2x².) By adding the two equations together, the x² terms cancel (or "eliminate") each other. Then all that's left is the relatively easy task of solving for x.

That is the physical point of intersection on the graph, determined simply by inspecting the lines drawn on the graph. Those lines intersect at the one point on the graph where x = +18 and y = -8.

If you're worried about the accuracy of the graphing, you could solve the simultaneous equations algebraically. See Solve Simultaneous Equations Using Elimination Method , Solve Simultaneous Equations Using Substitution Method and related articles.

Just select two or more small values of x (positive or negative), and then solve the equation for the corresponding values of y. For example, select +1 for the value of x, and solve the equation to see what y would be when x is +1. Do that with two (or more) x/y combinations for a linear (straight line) equation and three (or more) x/y combinations for a quadratic (curved-line) equation. You can choose any x values you want. Choosing small x values makes finding the corresponding y values easier.