Solving Systems of Linear Equations Graphically

a

To write the system of equations represented in the graph, we need to find the equations of the lines. Therefore, let's identify their slopes and

y

-intercepts.

The slopes of the lines are $m_{1} = \frac{- 3}{1} = - 3 and m_{2} = \frac{2}{1} = 2,$ and the $y$ -intercepts are $b_{1} = 1$ and $b_{2} = 6 .$ We can now write the system of equations by substituting the values into the slope-intercept form. ${y = - 3 x + 1 y = 2 x + 6$

b

The solution to this system of linear equations is the $x$ -value and $y$ -value value where the graphs intersect.

The solution is $(- 1, 4) .$

c

What does this actually mean? Well, it means that if we substitute

x = - 1

and

y = 4

into both equations in the system of equations, both equalities should hold true. Let's do this!

{y = 2 x + 6 y = - 3 x + 1 (I) (II)

SubstituteII

x = - 1

,

y = 4

{4 = ? 2 (- 1) + 6 4 = ? - 3 (- 1) + 1

MultPosNeg

a (- b) = - a \cdot b

{4 = ? - 2 + 6 4 = ? - 3 (- 1) + 1

MultNegNegOnePar

- a (- b) = a \cdot b

{4 = ? - 2 + 6 4 = ? 3 + 1

AddTermsAdd terms

{4 = 4 4 = 4

The solution is correct. In this particular case, you got

4 = 4

in both equations, however, that does not have to be the case. There could be two or more different solutions, what's important is that all of them are true.

Solving Systems of Linear Equations Graphically 1.6 - Solution