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Algebra · Simultaneous equations

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Simultaneous equations — one linear, one quadratic

Why a line can meet a curve twice (so expect two solution PAIRS), why substitution is the only route in, and the bracket-squaring discipline that circle equations demand.

Algebra · Simultaneous equations

Simultaneous equations — one linear, one quadratic

Why a line can meet a curve twice (so expect two solution PAIRS), why substitution is the only route in, and the bracket-squaring discipline that circle equations demand.

Why it works

A line and a parabola (or circle) can cross twice — so the algebra must be allowed to produce two answers. That's the headline difference from linear pairs: the solution is up to two pairs (x,y)(x, y), one per crossing point.

Substitution is forced. Elimination thrives on matching ax+byax + by shapes, but x2x^2 has no partner to cancel against. Instead, substitute the linear equation into the quadratic one. For y=x2−3x+4y = x^2 - 3x + 4 and y=x+1y = x + 1: both equal yy, so

x2−3x+4=x+1  ⇒  x2−4x+3=0  ⇒  (x−1)(x−3)=0.x^2 - 3x + 4 = x + 1 \;\Rightarrow\; x^2 - 4x + 3 = 0 \;\Rightarrow\; (x - 1)(x - 3) = 0.

x=1x = 1 or x=3x = 3 — and each xx needs its yy from the linear equation (it's simpler, and using the quadratic invites slips): (1,2)(1, 2) and (3,4)(3, 4).

Circles demand the full bracket square. Substituting y=x+1y = x + 1 into x2+y2=25x^2 + y^2 = 25:

x2+(x+1)2=25  ⇒  x2+x2+2x+1=25  ⇒  2x2+2x−24=0  ⇒  x2+x−12=0.x^2 + (x + 1)^2 = 25 \;\Rightarrow\; x^2 + x^2 + 2x + 1 = 25 \;\Rightarrow\; 2x^2 + 2x - 24 = 0 \;\Rightarrow\; x^2 + x - 12 = 0.

(x+1)2(x + 1)^2 is x2+2x+1x^2 + 2x + 1 — the middle term is where the marks live; x2+1x^2 + 1 term-by-term destroys the question. Factorising: (x+4)(x−3)=0(x + 4)(x - 3) = 0, so the crossing points are (3,4)(3, 4) and (−4,−3)(-4, -3).

Keep the pairs PAIRED. Each yy belongs to its own xx. Presenting "x=3x = 3 or −4-4, y=4y = 4 or −3-3" unmatched (or cross-matched) throws away the final mark — write coordinates, (3,4)(3, 4) and (−4,−3)(-4, -3), and check one pair in the ORIGINAL quadratic: 9+16=259 + 16 = 25. ✓

One repeated root = a tangent. If the quadratic collapses to a perfect square — (x−3)2=0(x - 3)^2 = 0 — the two crossings have merged: the line touches the curve at exactly one point. The algebra isn't broken; it's telling you the geometry.