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Algebra - Simultaneous Linear Equations in Two Variables

Grade 9ICSE

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

🔑Concepts

A Simultaneous Linear Equation in two variables consists of two equations of the form ax+by+c=0ax + by + c = 0, where xx and yy are variables and a,b,ca, b, c are real numbers such that a2+b20a^2 + b^2 \neq 0. These equations are 'simultaneous' because we seek a common pair of (x,y)(x, y) values that satisfy both equations at the same time.

The Method of Substitution involves expressing one variable in terms of the other from one equation and substituting this expression into the second equation. This transforms a two-variable system into a single linear equation in one variable, which is easier to solve.

The Method of Elimination involves multiplying one or both equations by suitable non-zero constants so that the coefficients of one variable become numerically equal (or opposites). By adding or subtracting the resulting equations, one variable is 'eliminated', allowing you to solve for the remaining variable.

The Method of Cross-Multiplication uses a specific ratio formula derived from the general forms a1x+b1y+c1=0a_1x + b_1y + c_1 = 0 and a2x+b2y+c2=0a_2x + b_2y + c_2 = 0. It follows a visual pattern where coefficients are arranged in columns to calculate xx and yy directly using products of diagonal terms.

Graphically, a linear equation in two variables represents a straight line on a Cartesian plane. When you plot two such equations, the point (x,y)(x, y) where the two lines intersect is the unique solution to the system. If the lines are parallel and never meet, the system has no solution.

A system of equations is 'Consistent' if it has at least one solution. Visually, this occurs when lines intersect at one point (Unique Solution) or coincide exactly (Infinite Solutions). It is 'Inconsistent' if there is no solution, which visually represents two parallel lines that never cross.

Solvability Conditions: For the equations a1x+b1y=c1a_1x + b_1y = c_1 and a2x+b2y=c2a_2x + b_2y = c_2, a unique solution exists only if a1a2b1b2\frac{a_1}{a_2} \neq \frac{b_1}{b_2}. If the ratios of coefficients of xx, yy, and the constants are all equal (a1a2=b1b2=c1c2\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}), the lines overlap completely, resulting in infinite solutions.

📐Formulae

Standard Form: a1x+b1y+c1=0a_1x + b_1y + c_1 = 0 and a2x+b2y+c2=0a_2x + b_2y + c_2 = 0

Condition for Unique Solution: a1a2b1b2\frac{a_1}{a_2} \neq \frac{b_1}{b_2}

Condition for No Solution (Parallel Lines): a1a2=b1b2c1c2\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}

Condition for Infinite Solutions (Coincident Lines): a1a2=b1b2=c1c2\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}

Cross-Multiplication Formula: xb1c2b2c1=yc1a2c2a1=1a1b2a2b1\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}

Solutions via Cross-Multiplication: x=b1c2b2c1a1b2a2b1x = \frac{b_1c_2 - b_2c_1}{a_1b_2 - a_2b_1} and y=c1a2c2a1a1b2a2b1y = \frac{c_1a_2 - c_2a_1}{a_1b_2 - a_2b_1}

💡Examples

Problem 1:

Solve the following system of equations using the Elimination Method: 3x+2y=113x + 2y = 11 2x+3y=42x + 3y = 4

Solution:

  1. Multiply the first equation by 33 and the second equation by 22 to make the coefficients of yy equal: 3(3x+2y)=3(11)    9x+6y=333(3x + 2y) = 3(11) \implies 9x + 6y = 33 (Eq 3) 2(2x+3y)=2(4)    4x+6y=82(2x + 3y) = 2(4) \implies 4x + 6y = 8 (Eq 4)
  2. Subtract Eq 4 from Eq 3: (9x4x)+(6y6y)=338(9x - 4x) + (6y - 6y) = 33 - 8 5x=255x = 25 x=5x = 5
  3. Substitute x=5x = 5 into the first original equation: 3(5)+2y=113(5) + 2y = 11 15+2y=1115 + 2y = 11 2y=11152y = 11 - 15 2y=4    y=22y = -4 \implies y = -2 Final Solution: x=5,y=2x = 5, y = -2.

Explanation:

We use the elimination method by equalizing the coefficients of yy. By subtracting the modified equations, we isolate xx. Once xx is found, we substitute it back into any original equation to find yy.

Problem 2:

Solve for xx and yy using the Substitution Method: x+y=7x + y = 7 5x+12y=75x + 12y = 7 (Wait, let's use 5x+12y=75x + 12y = 7 for variety)

Solution:

  1. From the first equation x+y=7x + y = 7, express xx in terms of yy: x=7yx = 7 - y
  2. Substitute x=7yx = 7 - y into the second equation: 5(7y)+12y=75(7 - y) + 12y = 7 355y+12y=735 - 5y + 12y = 7 35+7y=735 + 7y = 7 7y=7357y = 7 - 35 7y=287y = -28 y=4y = -4
  3. Substitute y=4y = -4 back into the expression for xx: x=7(4)x = 7 - (-4) x=7+4=11x = 7 + 4 = 11 Final Solution: x=11,y=4x = 11, y = -4.

Explanation:

We isolate xx from the simpler equation and plug that expression into the more complex equation. This reduces the problem to a single-variable equation, which we solve for yy before finding xx.