eg \(192 = 2a + 29(\text{``}{6}\text{''})\) oe or \(123 = a + 19(\text{``}{6}\text{''})\) oeor eg \(192 = 2(\text{``}{9}\text{''}) + 29d\) oe or \(123 = \text{``}{9}\text{''} + 19d\) oe
M1
Working required Answer: \(a = 9\) \(d = 6\)
A1
(5)
(5 marks)
Notes
M1: for using \(U_n = a + (n - 1)d\)
M1: for using \(S_n = \dfrac{n}{2}\left(2a + (n - 1)d\right)\)
M1: (dep on M2) for a correct method to find \(a\) or \(d\):
coefficients of \(a\) or \(d\) the same in correct equations and correct operator to eliminate selected variable resulting in an equation in \(a\) only or in \(d\) only
or
writing \(a\) or \(d\) in terms of the other variable and correctly substituting (condone missing brackets)
M1: (dep on M3) for substituting their found value of \(a\) or \(d\) into a correct equation
A1: dep on M2 \(a\) and \(d\) must be clearly identified
M1: for a correct expression or equation using the common difference, may be in terms of \(d\) for this mark we will allow an expression for \(-d\) or \(-2d\)
M1: for a correct equation for the sum of 9 terms in \(x\) and \(d\) or in terms of \(x\) and \(y\) or in terms of \(x\) or in terms of \(d\)
for “\(3y - 2x - 9\)” we will allow \((3y - 4) - (2x + 5)\) or for using their incorrect simplification from \((3y - 4) - (2x + 5)\) shown
for “\(4x - 3y + 2\)” we will allow \((4x - 2) - (3y - 4)\) or for using their incorrect simplification from \((4x - 2) - (3y - 4)\) shown
similarly for their “\(x - 3.5\)” and their “\(d + 3.5\)”
M2:left hand column 2 correct equations in terms of \(x\) and \(y\) in the form \(px + qy = r\) oe or 2 correct equations in terms of \(x\) and \(d\) in the form \(px + qd = r\) oe
right hand column 2 correct equations in terms of \(x\) and \(y\) where one is substituted into the other to get a correct equation in the form \(px + q = r\) or \(py + q = r\) or 2 correct equations in terms of \(x\) and \(d\) where one is substituted into the other to get a correct equation in the form \(px + q = r\) or \(pd + q = r\)
If not M2 then M1 for one correct equation in any of the required forms from the left hand or right hand column
A2: (dep on M2) oe (allow 7.3(33…))
(A1 for \(x = \dfrac{11}{2}\) or \(y = \dfrac{22}{3}\))
Answer: \(x = -\dfrac{1}{2}\) and \(y = \dfrac{3}{2}\)
\(x = -3\) and \(y = -1\)
A1
(5)
(5 marks)
Notes
M1: for substitution of \(y = x + 2\) (or \(x = \pm y \pm 2\)) into \(x^2 + 3y + y^2 = 7\) to obtain an equation in \(x\) only (or \(y\) only)
M1ft: dep on previous M1 for multiplying out and collecting terms, forming a three term quadratic in any form of \(ax^2 + bx + c\;(= 0)\) where at least 2 coefficients (\(a\) or \(b\) or \(c\)) are correct
M1ft: dep on first M1 method to solve their 3 term quadratic using any correct method (allow one sign error and some simplification – allow as far as eg \(\dfrac{-7 \pm \sqrt{49 - 24}}{4}\) or \(\dfrac{1 \pm \sqrt{1 + 24}}{4}\)
or if factorising allow brackets which expanded give 2 out of 3 terms correct)
or correct values for \(x\)
or correct values for \(y\)
M1ft: dep on previous M1 for substituting their 2 found values of \(x\) or \(y\) into one of the two given equations or fully correct values for the other variable (correct labels for \(x\) / \(y\)) or for one correct pair of values
A1: oe dep on M2 (allow coordinates)
If they find the values of \(y\) but think they are the values of \(x\) then the maximum mark is 3
M1: substitution of \(y = \pm 2x \pm 3\) (or \(x = \dfrac{\pm y \pm 3}{2}\)) into \(x^2 + y^2 = 41\) to obtain an equation in \(x\) only (or \(y\) only)
M1ft: dep on previous M1 for multiplying out and collecting terms, forming a three term quadratic in any form of \(ax^2 + bx + c\;(= 0)\) where at least 2 coefficients (\(a\) or \(b\) or \(c\)) are correct
M1ft: dep on M1 method to solve their 3 term quadratic using any correct method (allow one sign error and some simplification – allow as far as eg \(\dfrac{12 \pm \sqrt{144 + 640}}{10}\) or \(\dfrac{6 \pm \sqrt{36 + 3100}}{10}\) or if factorising allow brackets which expanded give 2 out of 3 terms correct) or correct values for \(x\) or correct values for \(y\)
M1ft: dep on previous M1 for substituting their 2 found values of \(x\) or \(y\) into one of the two given equations or their rearranged equation used in the substitution or for one correct pair of values
A1: oe dep on M2 for all 4 values (allow coordinates)
If they find the values of \(y\) but think they are the values of \(x\) then the maximum mark is 3
M1: for a correct method to eliminate \(x\) or \(y\): coefficients of \(x\) or \(y\) the same and correct operator to eliminate selected variable (condone any one arithmetic error in multiplication) or writing \(x\) or \(y\) in terms of the other variable and correctly substituting (condone missing brackets)
NB The mark is for the method and not for the result of the method. However, if the correct result of the method is seen, the mark can be awarded.
M1: dep on first M1 for a correct method to find other variable by substitution of found variable into one equation or for repeating the above method to find the second variable.
M1: a correct method to eliminate \(x\) or \(y\): coefficients of \(x\) or \(y\) are the same and the correct operation to eliminate is selected; if operator not written, the correct operation can be implied by 2 out of 3 terms correct Allow one arithmetic error if multiplying to equate coefficients or for a correct substitution of one variable into the other equation
NB: the mark is for the method and not for the result of the method. However, if the correct result of this method is seen, the mark can be awarded
M1: dep on M1 a correct substitution to find the value of the second variable using their value or for starting again with elimination or substitution (as above)
M1: substitution of linear equation into quadratic – allow one sign error in substituted expression
M1: Dep on M1 simplified to a 3 term quadratic with 2 or 3 of 3 terms correct
M1ft: dep on M1 for solving their 3 term quadratic equation using any correct method (if factorising, allow brackets which expanded give 2 out of 3 terms correct ) (if using formula allow one sign error and some simplification – allow as far as \(\dfrac{3 \pm \sqrt{9 - 8}}{4}\) or \(\dfrac{-5 \pm \sqrt{25 - 24}}{4}\) ) or if completing the square then as far as shown on LHS OR the correct values for \(x\) OR the correct values for \(y\)
M1: dep on previous M1 for correct method to find both other values or a correct pair of values
