AC9M10A02Year 10AlgebraPractised on site

AC9M10A02 — Year 10 Algebra

Official Curriculum Content Description:

solve linear inequalities and simultaneous linear equations in 2 variables; interpret solutions graphically and communicate solutions in terms of the situation

Worked Examples (3)

Example 1Solving Simultaneous Linear Equations
Solve the simultaneous equations by substitution: −x−3y=20-x - 3y = 20 −3x−3y=30-3x - 3y = 30
Correct Answer:(−5,−5)(-5, -5)
Worked Solution:
Use substitution to solve the system: Equation 1: −x−3y=20-x - 3y = 20 Equation 2: −3x−3y=30-3x - 3y = 30 Rearrange one equation to express a variable, then substitute into the other. This yields x=−5x = -5 and y=−5y = -5. Written as a coordinate pair: (−5,−5)(-5, -5).
Example 2Solving Linear Inequalities
Solve the inequality x−9<−10x - 9 < -10.
Correct Answer:x<−1x < -1
Worked Solution:
Subtract -9 from both sides: x<−1x < -1 Divide both sides by 1: x<−1x < -1.
Example 3Equation of a Straight Line (y = mx + c)
Find the gradient of the line passing through the points (−4,19)(-4, 19) and (−2,11)(-2, 11).
Common mistake: Subtracting the coordinates in an inconsistent order, giving the negative of the correct gradient
Correct Answer:−4-4
Worked Solution:
Use the gradient formula m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}: m=11−(19)−2−(−4)=−82=−4m = \frac{11 - (19)}{-2 - (-4)} = \frac{-8}{2} = -4.

Ready to practise this descriptor?

Work through unlimited interactive questions for this curriculum skill, with instant answers and step-by-step methods.

Start unlimited online practicePrint Year 10 worksheet