Solving one linear and one quadratic equation simultaneously by substitution

Reading the solutions as the intersection points of a line and a curve

Using the discriminant to decide whether a line crosses, touches or misses a curve

Linear inequalities, including ones with brackets and fractions

Quadratic inequalities, including ones that need expanding first

Inequalities in a parameter such as k that come from a discriminant condition

Combining inequalities and writing solution sets in set notation

Regions on a graph: solid and dashed boundaries, and describing a region with inequalities

1. To solve a linear and a quadratic equation together, make one variable the subject of the linear equation, substitute into the quadratic, solve the resulting quadratic, and then use the linear equation to find the partner of each value.

2. Each solution pair is a point where the line meets the curve; after substituting, the discriminant b²-4ac of the quadratic tells you how many: positive gives two points, zero means the line is a tangent, negative means no points.

3. Linear inequalities are solved like equations, except that multiplying or dividing by a negative number reverses the inequality sign.

4. Quadratic inequalities: rearrange to f(x)>0 or f(x)<0, find the critical values (the roots), sketch, and read off where the curve is above or below the x-axis.

5. Set notation: {x: x<-4}∪{x: x>2} for "or"; {x: -(3/2)≤x<1} for a single interval.

6. Regions: a solid boundary is included (≤, ≥); a dashed boundary is not (<, >).

With two linear equations you can often add or subtract to eliminate a variable. When one equation contains x², y² or xy, elimination usually fails: the squared and product terms do not cancel. The method that always works is substitution. The linear equation gives one variable in terms of the other with no squares at all, so you can replace that variable everywhere in the quadratic and be left with a single equation in one unknown.

The substitution method

1. Rearrange the linear equation to make x or y the subject. Choose the one that avoids fractions: if a variable has coefficient 1, use it.

2. Substitute into the quadratic equation and expand carefully. Put brackets round the whole substituted expression before squaring or multiplying.

3. Solve the resulting quadratic in one variable: factorise, complete the square or use the formula.

4. Find the partner of each value by substituting into the linear equation. The linear equation gives exactly one partner each time; the quadratic might give two, and one of them would be false.

5. Pair the answers: write them as coordinates or as matched statements x=…, y=….

If both equations already have y as the subject, step 1 is free: set the two right-hand sides equal to each other.

WORKED EXAMPLE · QUESTION

Solve the simultaneous equations y=x²-4x+5 and y=x+1.

SOLUTION

Both expressions equal y, so set them equal: x²-4x+5=x+1, which gives x²-5x+4=0.

Factorise: (x-1)(x-4)=0, so x=1 or x=4.

Use the linear equation y=x+1 for the partners: x=1 gives y=2, and x=4 gives y=5.

Check in the quadratic: 1-4+5=2 ✓ and 16-16+5=5 ✓. x=1, y=2 and x=4, y=5

WORKED EXAMPLE · QUESTION

Solve the simultaneous equations x+3y=4 and x²+xy+y²=3.

SOLUTION

Choose the subject. x has coefficient 1 in the linear equation, so x=4-3y (no fractions that way).

Substitute, with brackets, and expand. (4-3y)²+(4-3y)y+y²=3, so 16-24y+9y²+4y-3y²+y²=3, which simplifies to 7y²-20y+13=0.

Solve. (y-1)(7y-13)=0, so y=1 or y=(13/7).

Partners from x=4-3y. y=1 gives x=1; y=(13/7) gives x=4-(39/7)=-(11/7).

x=1, y=1 and x=-(11/7), y=(13/7)

Every real solution pair is a point that lies on both graphs, so it is a point where the line meets the curve. After you substitute, everything about the intersection is stored in one quadratic equation. Its number of real roots is the number of intersection points, and the discriminant b²-4ac of that quadratic counts the roots without solving it.

This is the key idea of the whole page. A geometric question ("does the line touch the curve?") becomes an algebraic one ("is this discriminant zero?"). The idea comes back with circles in P2 and with tangents in calculus.

WORKED EXAMPLE · QUESTION

The curve C has equation y=x²+x-1.

(a) Show that the line y=3x-2 is a tangent to C, and find the point of contact.

(b) Find the set of values of c for which the line y=3x+c meets C at two distinct points.

SOLUTION

(a) Substitute: x²+x-1=3x-2, so x²-2x+1=0. For a tangent we need b²-4ac=0: here (-2)²-4(1)(1)=0, so there is a repeated root and the line is a tangent.

Point of contact: (x-1)²=0 gives x=1, and then y=3(1)-2=1. Point of contact (1, 1)

(b) Substitute the general line: x²+x-1=3x+c, so x²-2x-(1+c)=0. Two distinct points means discriminant >0:

(-2)²-4(1)(-(1+c))=4+4(1+c)=8+4c>0, so c>-2. {c: c>-2}

This is consistent with (a): c=-2 gives the tangent itself, the boundary case.

An inequality such as 3x-1<8 describes a whole range of values, not a single number. You solve it with the same steps as a linear equation: expand brackets, collect terms, divide. Almost every step keeps the inequality sign exactly as it was:

Adding or subtracting the same quantity on both sides never changes the sign.

Multiplying or dividing by a positive number keeps the sign.

Multiplying or dividing by a negative number reverses it: -2x<6 gives x>-3.

Why the reversal? Multiplying by -1 reflects the number line about 0, so the order of every pair of numbers flips: 2<5 but -2>-5. With fractions, multiply every term by the lowest common denominator first; it is a positive number, so the sign is safe. You can avoid dividing by a negative altogether by collecting the x terms on the side where their coefficient ends up positive.

WORKED EXAMPLE · QUESTION

Find the set of values of x for which (a) 5-3(x-1)>2x-7 (b) (2x-1)/3-(x+4)/4≤x-2.

SOLUTION

(a) Expand: 5-3x+3>2x-7, so 8-3x>2x-7. Collect the x terms on the right, where they are positive: 15>5x. Divide by 5: x<3

(b) Multiply every term by 12: 4(2x-1)-3(x+4)≤12x-24, so 8x-4-3x-12≤12x-24, giving 5x-16≤12x-24.

Collect on the right and divide by the positive number 7: 8≤7x. {x: x≥(8/7)}

You cannot solve x²+2x-8>0 by "moving things across" in the same way, because the sign of a quadratic changes at its roots. The reliable method uses a sketch.

Four steps

1. Rearrange so that one side is zero: f(x)>0, <0, ≥0 or ≤0. Expand any brackets first.

2. Find the critical values: solve f(x)=0.

3. Sketch y=f(x), marking the roots on the x-axis.

4. Read off where the curve is above the axis (f(x)>0) or below it (f(x)<0).

For a positive x² coefficient and roots α<β, the curve is a ∪-shape, so:

f(x)<0 for α<x<β: one interval, between the roots;

f(x)>0 for x<α or x>β: two separate intervals, outside the roots.

For "≤" or "≥" the critical values themselves are included, because there f(x)=0.

WORKED EXAMPLE · QUESTION

Solve the inequality (x-1)(x+5)>2x+3.

SOLUTION

Expand and rearrange to compare with zero: x²+4x-5>2x+3, so x²+2x-8>0.

Critical values: (x+4)(x-2)=0, so x=-4 or x=2.

The graph is ∪-shaped, and we want the parts above the axis, which are outside the roots. x<-4 or x>2

If the x² coefficient is negative, either sketch the ∩-shape directly or multiply through by -1 first and reverse the sign. For example, 6+x-x²>0 becomes x²-x-6<0, that is (x-3)(x+2)<0, so -2<x<3. Both routes give the same answer, as the sketch shows.

Discriminant questions often end in a quadratic inequality, not in x but in a constant such as k. The condition "the line misses the curve" means b²-4ac<0, and b or c contains k, so the discriminant is itself a quadratic in k. Solve it in exactly the same way: critical values, then a sketch, now with k on the horizontal axis.

WORKED EXAMPLE · QUESTION

The line y=kx+1, where k is a constant, does not meet the curve y=x²+2x+5. Find the set of possible values of k.

SOLUTION

Substitute and collect into a quadratic in x: x²+2x+5=kx+1, so x²+(2-k)x+4=0.

No intersection means no real roots, so b²-4ac<0: (2-k)²-4(1)(4)<0, which expands to k²-4k-12<0.

Critical values: (k-6)(k+2)=0, so k=-2 or k=6.

"<0" on a ∪-shaped graph is between the roots. {k: -2<k<6}

At the critical values the line is a tangent: k=-2 touches the curve at (-2, 5) and k=6 touches it at (2, 13).

When x must satisfy two inequalities at once, the answer is the overlap of the two solution sets. The safest way to find it is to draw both on number lines stacked above each other and read off where both are shaded. A filled circle marks an end value that is included (≤, ≥); an open circle one that is not (<, >).

An answer that is one interval, such as -(3/2)≤x<1, can be written as a single double inequality. An answer in two separate pieces needs the word "or", or a union symbol in set notation. The braces {x: …} are read "the set of all x such that…".

WORKED EXAMPLE · QUESTION

Find the set of values of x for which both 4(x+2)<3x+9 and 2x²-3x-9≤0. Give your answer in set notation.

SOLUTION

Linear inequality. 4x+8<3x+9, so x<1.

Quadratic inequality. Critical values: (2x+3)(x-3)=0, so x=-(3/2) or x=3. "≤0" on a ∪-shape is between the roots, with the roots included: -(3/2)≤x≤3.

Overlap. The lower end comes from the quadratic and is included; the upper end comes from the linear inequality and is excluded. {x: -(3/2)≤x<1}

An inequality in x and y, such as y>x²-4, is satisfied by a whole region of the plane. To draw it:

Draw the boundary by replacing the inequality sign with "=".

Use a solid boundary for ≤ or ≥ (points on it are included) and a dashed boundary for < or > (not included).

y>f(x) is the region above the curve y=f(x); y<f(x) is below it.

If you are unsure which side, test a point that is not on the boundary; the origin is often easiest.

Shade and label the region you are asked for, unless the question gives a different instruction.

The same picture answers one-variable questions too: f(x)<g(x) holds for the x-values where the graph of f is below the graph of g, which is either between or outside their intersection points.

WORKED EXAMPLE · QUESTION

The curve C has equation y=x²-2x-3 and the line L has equation y=x+1.

(a) Find the coordinates of the points where L meets C.

(b) The region R lies above C and below L. Points on C are included in R; points on L are not. Write down inequalities that define R.

(c) The line y=x+k is a tangent to C. Find the value of k and the coordinates of the point of contact.

(d) Hence, or otherwise, find the set of values of x for which x²-2x-3<x+1.

SOLUTION

(a) x²-2x-3=x+1, so x²-3x-4=0, so (x+1)(x-4)=0. Then y=x+1 gives (-1, 0) and (4, 5)

(b) Above C with the curve included (≤); below L with the line excluded (<): x²-2x-3≤y<x+1

(c) Substitute y=x+k: x²-3x-(3+k)=0. Tangent means discriminant =0: 9+4(3+k)=0, so 21+4k=0 and k=-(21/4).

The repeated root is x=-b/2a=(3/2), and y=(3/2)-(21/4)=-(15/4). Check on the curve: (9/4)-3-3=-(15/4) ✓.

k=-(21/4), point of contact ((3/2), -(15/4))

(d) The curve is below the line between the intersection points from (a). {x: -1<x<4}

P1.1 Indices, surds and quadratics

Factorising, completing the square and b²-4ac are the tools used at every step here; this page applies them to two equations at once.

P1.3 Curve sketching and P1.4 Straight-line graphs

Sketches show how many times graphs meet, and perpendicular lines and intersection points use the same substitution method.

Circles (P2) and linear programming (D1)

A line meets a circle at two, one or no points, decided by exactly the same discriminant test; a tangent to a circle is the case b²-4ac=0. In Decision Mathematics, a set of linear inequalities defines a feasible region on a graph, with the same boundary and shading conventions.

1. Solve the simultaneous equations y=2x and x²+y²=20.

2. Solve 3-2x≥11.

3. Solve x²<9x.

4. Find the value of k for which y=x+k is a tangent to y=x²-3x+1, and the point of contact.

5. Which of the points (1, -2) and (3, -1) lie in the region defined by y≥x²-4 and y<0?

6. Find the set of values of p for which the equation x²+px+p+3=0 has two distinct real roots.

1. Substitute y=2x: x²+4x²=20, so x²=4 and x=±2. Using y=2x for the partners: (2, 4) and (-2, -4).

2. -2x≥8. Dividing by -2 reverses the sign: x≤-4.

3. x²-9x<0, so x(x-9)<0. Critical values 0 and 9; the ∪-shaped curve is below the axis between them: 0<x<9. (Dividing by x would have lost the lower limit.)

4. x²-3x+1=x+k, so x²-4x+1-k=0. Tangent: 16-4(1-k)=0, so 12+4k=0 and k=-3. Then x²-4x+4=0, (x-2)²=0, x=2, y=2-3=-1. k=-3, point of contact (2, -1).

5. (1, -2): -2≥1-4=-3 ✓ and -2<0 ✓, so it is inside. (3, -1): -1≥9-4=5 is false, so it is outside. Only (1, -2).

6. Two distinct roots: p²-4(p+3)>0, so p²-4p-12>0, (p-6)(p+2)>0. Critical values -2 and 6; ">0" on a ∪-shape is outside the roots: {p: p<-2}∪{p: p>6}.