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Line 1 Equation1x + 1y = 3
Line 2 Equation1x + -1y = 1
Intersection Solution (2.0, 1.0)

Variable Adjuster

Line 1: x coefficient (a1)1
-55
Line 1: y coefficient (b1)1
-55
Line 1: constant (c1)3
-1010
Line 2: x coefficient (a2)1
-55
Line 2: y coefficient (b2)-1
-55
Line 2: constant (c2)1
-1010

Systems of Linear Equations

SYS

A system of linear equations consists of two or more equations. Graphically, the solution to the system is the exact intersection point where the lines cross, representing the coordinates that satisfy both equations simultaneously.

a1x+b1y=c1anda2x+b2y=c2a_1x + b_1y = c_1 \quad \text{and} \quad a_2x + b_2y = c_2

Whiteboard Solver Steps

Step 1

Identify Equations (Lines on the Grid)

We begin with two linear equations. Geometrically, each equation represents a straight line on a 2D Cartesian plane. Solving the system means finding the coordinates (x, y) where the two lines intersect.

Line 1: 1x+1y=3Line 2: 1x+−1y=1\begin{aligned}\text{Line 1: } 1x + 1y = 3 \\ \text{Line 2: } 1x + -1y = 1\end{aligned}
Step 2

Calculate Main Determinant (D)

We calculate the main determinant D of the coefficient matrix. Since D is not equal to zero (D=D = {D}),weconfirmthatthetwolinesarenotparallelandhaveasingle,uniqueintersectionpoint.), we confirm that the two lines are not parallel and have a single, unique intersection point.

D=∣a1b1a2b2∣=a1b2−a2b1=(1)(−1)−(1)(1)=−2\begin{aligned}D = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} = a_1b_2 - a_2b_1 = (1)(-1) - (1)(1) = -2\end{aligned}
Step 3

Calculate Variable Determinants (Dx and Dy)

We calculate Dx (by replacing the x-coefficient column with the constant vector c) and Dy (by replacing the y-coefficient column with c). These determinants represent the scaling components along each coordinate axis.

Dx=∣c1b1c2b2∣=c1b2−c2b1=(3)(−1)−(1)(1)=−4Dy=∣a1c1a2c2∣=a1c2−a2c1=(1)(1)−(1)(3)=−2\begin{aligned}D_x = \begin{vmatrix} c_1 & b_1 \\ c_2 & b_2 \end{vmatrix} = c_1b_2 - c_2b_1 = (3)(-1) - (1)(1) = -4 \\ D_y = \begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix} = a_1c_2 - a_2c_1 = (1)(1) - (1)(3) = -2\end{aligned}
Step 4

Solve for Intersection Coordinates

Using Cramer's Rule, we divide the variable determinants by the main determinant D. The coordinates of the intersection point are: x = 2 and y = 1. At this unique point, both linear equations are satisfied simultaneously.

x=DxD=−4−2=2y=DyD=−2−2=1\begin{aligned}x = \frac{D_x}{D} = \frac{-4}{-2} = 2 \\ y = \frac{D_y}{D} = \frac{-2}{-2} = 1\end{aligned}