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Solving two equations in two unknowns by elimination or substitution, including one linear and one quadratic equation. The solutions are the points where the graphs intersect, and substituting a line into a curve and using the discriminant tells you whether they meet at all. Typical questions are worth 4 to 7 marks.
Key facts
Elimination: same signs subtract, different signs add
One linear, one quadratic: use substitution
Substitute back into the linear equation
Solutions = coordinates of the intersection points
After substituting: \(b^2 - 4ac \gt 0\) two points, \(= 0\) tangent, \(\lt 0\) no intersection
Elimination: multiply one or both equations so that the coefficients of \(x\) (or \(y\)) are the same size; subtract if the signs are the same, add if they are different.
Substitution: make \(x\) or \(y\) the subject of one equation and substitute it into the other.
Check your answer in the equation you did not use last.
One linear and one quadratic
Always use substitution. Rearrange the linear equation, choosing the variable that avoids fractions.
Substitute into the quadratic, expand, collect every term on one side and solve the quadratic.
Substitute each value back into the linear equation to find the matching value. Solutions come in pairs.
Example: \(y = x + 1\) and \(x^2 + y^2 = 13\) give \(x^2 + (x + 1)^2 = 13\), so \(2x^2 + 2x - 12 = 0\), i.e. \(x^2 + x - 6 = 0\). So \(x = 2, y = 3\) or \(x = -3, y = -2\)
What the solutions mean
diagram
The solutions are the coordinates of the points where the two graphs intersect.
Each solution pair is a point where the line meets the curve: here (2, 3) and (−3, −2).
After substituting a line into a curve you get a quadratic. Its discriminant tells you how they meet:
\(b^2 - 4ac \gt 0\): two points of intersection; \(b^2 - 4ac = 0\): the line is a tangent (one point); \(b^2 - 4ac \lt 0\): the line and curve do not meet.
Curves with \(xy\) or \(y^2\) terms work the same way; just expand every bracket carefully after substituting. Grade A or above only
How to answer each type of question
M = method mark, A = answer mark, B = a mark on its own.
Find constants from two conditions
3 marksGrade D
Substitute each given point into the equation.
Solve the two linear equations simultaneously.
Check both points in your final equation.
Example. The curve \(y = ax^2 + bx\) passes through the points \((1, 4)\) and \((2, 2)\). Find the values of \(a\) and \(b\)
Show the model answerHide the model answer
\(a + b = 4\) and \(4a + 2b = 2\)
(M1)
\(2a + b = 1\), so subtracting: \(a = -3\)
(M1)
\(a = -3\), \(b = 7\)
(A1)
Solve one linear and one quadratic equation
6 marksGrade C
Rearrange the linear equation for one variable.
Substitute into the quadratic and simplify to a three-term quadratic equal to 0.
Solve it.
Substitute each value into the linear equation and pair the answers.
Example. In this question you must show all stages of your working. Solve the simultaneous equations \(x + 2y = 5\) \(x^2 + 2y^2 = 11\)
Answer: \(x = -5, \ y = -21\) and \(x = 2, \ y = 7\)
M1 for substituting the linear equation into the quadratic: \(x^{2} + 7x - 11 = 4x - 1\)
A1 for a correct three-term quadratic: \(x^{2} + 3x - 10 = 0\)
dM1 for solving their quadratic
A1 for \(x = -5\) and \(x = 2\)
A1 for \(y = -21\) and \(y = 7\), correctly paired
Worked solution: \(x^{2} + 7x - 11 = 4x - 1\) \(x^{2} + 3x - 10 = 0\) \((x + 5)(x - 2) = 0\), so \(x = -5\) or \(x = 2\). Using \(y = 4x - 1\): when \(x = -5\), \(y = -21\); when \(x = 2\), \(y = 7\).
Question 2Medium6 marks
In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
Solve the simultaneous equations \(3x + y = 5\) \(x^2 + 4xy + y^2 = 33\)[6]
Show the answer and mark scheme
Answer: \(x = -4, \ y = 17\) and \(x = -1, \ y = 8\)
M1 for substituting \(y = -3x + 5\) into the second equation
M1 for expanding the brackets correctly (at least two of the three terms in \(x\) correct)
A1 for a correct three-term quadratic, e.g. \(x^{2} + 5x + 4 = 0\)
dM1 for solving their quadratic
A1 for \(x = -4\) and \(x = -1\)
A1 for \(y = 17\) and \(y = 8\), correctly paired
Worked solution: Substitute \(y = -3x + 5\): \(x^2 + 4x(-3x + 5) + (-3x + 5)^2 = 33\) \(-2x^{2} - 10x - 8 = 0\), i.e. \(x^{2} + 5x + 4 = 0\) \((x + 4)(x + 1) = 0\), so \(x = -4\) or \(x = -1\). Then \(y = 17\) and \(y = 8\) respectively.
Question 3Hard8 marks
In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. The circle \(C\) has equation \((x - 3)^2 + (y + 3)^2 = 10\). The line \(l\) has equation \(y = 3x + k\), where \(k\) is a constant.
(a) Show that the \(x\)-coordinates of any points of intersection of \(l\) and \(C\) satisfy \(10x^2 + (6k + 12)x + (k^2 + 6k + 8) = 0\)[3]
(b) Given that \(l\) is a tangent to \(C\), find the possible values of \(k\).[5]
Show the answer and mark scheme
(a)Answer: Substitute \(y = 3x + k\) into the equation of \(C\), expand and collect terms.
M1 for substituting \(y = 3x + k\) into the equation of \(C\)
M1 for expanding, e.g. \((3x + k + 3)^2 = 9x^2 + 6(k + 3)x + (k + 3)^2\)
A1* for collecting terms to obtain the given equation with no errors