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Linear Equations in Two Variables - Solution of a Linear Equation

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

A linear equation in two variables is an equation that can be written in the form ax+by+c=0ax + by + c = 0, where a,b,a, b, and cc are real numbers and aa and bb are not both zero. Visually, this equation represents a perfectly straight line when plotted on a Cartesian coordinate system.

A 'solution' to such an equation is a pair of values (x,y)(x, y), written as an ordered pair, which satisfies the equation by making the Left Hand Side (LHS) equal to the Right Hand Side (RHS). Geometrically, every such solution corresponds to a specific point located on the line representing the equation.

A linear equation in two variables has infinitely many solutions. You can visualize this as a line that extends infinitely in both directions; every single point on that line represents a unique pair of (x,y)(x, y) that solves the equation.

To find solutions for an equation, you can substitute any value for one variable (usually xx) and solve the resulting linear equation in one variable to find the corresponding value of yy. This process creates a collection of points that can be plotted to visualize the path of the line.

The points where the line crosses the axes are special solutions. The yy-intercept occurs where the line crosses the vertical axis (found by setting x=0x = 0), and the xx-intercept occurs where the line crosses the horizontal axis (found by setting y=0y = 0).

If a point (p,q)(p, q) lies on the line, then x=p,y=qx = p, y = q is a solution of the equation. Conversely, if a point does not lie on the line, the coordinates of that point will not satisfy the equation, meaning the point is visually 'off' the line's path.

The graph of every linear equation in two variables is a straight line. Because only two distinct points are needed to define a straight line, finding just two solutions is enough to draw the entire graph, though finding a third point is recommended to check for accuracy.

📐Formulae

Standard Form: ax+by+c=0ax + by + c = 0

Slope-Intercept Form: y=mx+cy = mx + c

Isolated variable form for calculation: y=frac(ax+c)by = \\frac{-(ax + c)}{b}

Equation of xx-axis: y=0y = 0

Equation of yy-axis: x=0x = 0

💡Examples

Problem 1:

Find three different solutions for the equation 2x+3y=122x + 3y = 12.

Solution:

Step 1: Let x=0x = 0. Substituting into the equation: 2(0)+3y=12Rightarrow3y=12Rightarrowy=42(0) + 3y = 12 \\Rightarrow 3y = 12 \\Rightarrow y = 4. So, (0,4)(0, 4) is a solution. \nStep 2: Let y=0y = 0. Substituting into the equation: 2x+3(0)=12Rightarrow2x=12Rightarrowx=62x + 3(0) = 12 \\Rightarrow 2x = 12 \\Rightarrow x = 6. So, (6,0)(6, 0) is a solution. \nStep 3: Let x=3x = 3. Substituting into the equation: 2(3)+3y=12Rightarrow6+3y=12Rightarrow3y=6Rightarrowy=22(3) + 3y = 12 \\Rightarrow 6 + 3y = 12 \\Rightarrow 3y = 6 \\Rightarrow y = 2. So, (3,2)(3, 2) is a solution.

Explanation:

To find solutions, we arbitrarily chose values for one variable and solved for the other. Each pair (x,y)(x, y) represents a point on the graph of the line 2x+3y=122x + 3y = 12.

Problem 2:

Check if the point (2,2)(2, -2) is a solution to the equation x2y=6x - 2y = 6.

Solution:

Step 1: Identify the coordinates x=2x = 2 and y=2y = -2. \nStep 2: Substitute these values into the LHS of the equation: LHS=x2y=22(2)LHS = x - 2y = 2 - 2(-2). \nStep 3: Simplify the expression: 2+4=62 + 4 = 6. \nStep 4: Compare LHS and RHS. Since LHS=6LHS = 6 and RHS=6RHS = 6, LHS=RHSLHS = RHS.

Explanation:

Because the substitution makes the equation true, the ordered pair (2,2)(2, -2) is a solution, meaning this point lies exactly on the line x2y=6x - 2y = 6.