June 2016 Paper 2 Q4
4 Work out the point of intersection of the lines
\[2x + 3y = 11 \qquad \text{and} \qquad 2y = 13 - 3x\][4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 Equates coefficients and eliminates a variable | ||
| \(4x + 6y = 22\) and \(9x + 6y = 39\) or \(6x + 9y = 33\) and \(6x + 4y = 26\) | M1 | oe Equates coefficients of one variable Allow one term error |
| \(4x - 9x =\) \(22 - 39\) or \(-5x = -17\) or \(9y - 4y =\) \(33 - 26\) or \(5y = 7\) | M1dep | oe Eliminates a variable Must be correct method for their equations Unless correctly eliminated, intention to subtract (or add if appropriate) must be seen with the result of their subtraction |
| (\(x\) =) 3.4 or (\(y\) =) 1.4 | A1 | oe eg \(x = \dfrac{17}{5}\) |
| (3.4, 1.4) | A1 | oe eg \(\left(\dfrac{17}{5}, \dfrac{7}{5}\right)\) SC3 (1.4, 3.4) oe |
| Alternative method 2 Makes a variable the subject in first equation | ||
| \(y = \dfrac{11}{3} - \dfrac{2}{3}x\) or \(x = 5.5 - 1.5y\) | M1 | oe eg \(x = \dfrac{11 - 3y}{2}\) Makes \(y\) or \(x\) the subject Allow one term error |
| \(2\left(\dfrac{11}{3} - \dfrac{2}{3}x\right) = 13 - 3x\) or \(2y = 13 - 3(5.5 - 1.5y)\) | M1dep | oe Eliminates a variable Must be correct method for their equations |
| (\(x\) =) 3.4 or (\(y\) =) 1.4 | A1 | oe eg \(x = \dfrac{17}{5}\) |
| (3.4, 1.4) | A1 | oe eg \(\left(\dfrac{17}{5}, \dfrac{7}{5}\right)\) SC3 (1.4, 3.4) oe |
| Alternative method 3 Makes a variable the subject in second equation | ||
| \(y = 6.5 - 1.5x\) or \(x = \dfrac{13}{3} - \dfrac{2}{3}y\) | M1 | oe eg \(y = \dfrac{13 - 3x}{2}\) Makes \(y\) or \(x\) the subject Allow one term error |
| \(2x + 3(6.5 - 1.5x) = 11\) or \(2\left(\dfrac{13}{3} - \dfrac{2}{3}y\right) + 3y = 11\) | M1dep | oe Eliminates a variable Must be correct method for their equations |
| (\(x\) =) 3.4 or (\(y\) =) 1.4 | A1 | oe eg \(x = \dfrac{17}{5}\) |
| (3.4, 1.4) | A1 | oe eg \(\left(\dfrac{17}{5}, \dfrac{7}{5}\right)\) SC3 (1.4, 3.4) oe |
| Alternative method 4 Makes same variable the subject in both equations | ||
| \(y = \dfrac{11}{3} - \dfrac{2}{3}x\) or \(y = 6.5 - 1.5x\) or \(x = 5.5 - 1.5y\) or \(x = \dfrac{13}{3} - \dfrac{2}{3}y\) | M1 | oe eg \(y = \dfrac{13 - 3x}{2}\) Makes \(y\) or \(x\) the subject Allow one term error |
| \(\dfrac{11}{3} - \dfrac{2}{3}x = 6.5 - 1.5x\) or \(5.5 - 1.5y = \dfrac{13}{3} - \dfrac{2}{3}y\) | M1dep | oe Eliminates a variable Must be correct method for their equations |
| (\(x\) =) 3.4 or (\(y\) =) 1.4 | A1 | oe eg \(x = \dfrac{17}{5}\) |
| (3.4, 1.4) | A1 | oe eg \(\left(\dfrac{17}{5}, \dfrac{7}{5}\right)\) SC3 (1.4, 3.4) oe |
Additional guidance
| Answer only (3.4, 1.4) | M2 A2 |
| One value correct (possibly by drawing) with no incorrect working seen for that variable | M2 A1 A0 |
| If the same method is used for both \(x\) and \(y\) (eg equates coefficients and eliminates a variable), mark the attempt that favours the student | |
| Alt 1 \(\quad \begin{aligned} 6x + 9y &= 33 \\ \underline{6x + 4y} &\underline{{}= 26} \\ 4y &= 6 \end{aligned}\) is M1 M0 unless intention to subtract is seen (eg a subtraction symbol is seen or the word subtract is seen) which would then get M1 M1 | |
| Alts 2, 3 and 4 Allow rounding or truncating to 1dp or better for up to M1 M1 eg (Alt 2) \(\ y = 3.6 - 0.6x\) \(2(3.6 - 0.6x) = 13 - 3x\) | M1 M1 |
| (1.4, 3.4) is SC3 or M2 A2 if \(x = 3.4\) and \(y = 1.4\) seen in working |