Simultaneous equations – Year 10 GCSE maths revision notes to download as PDF

Worked example 1

\(x+y=10\) and \(x-y=4\).

Add: \(2x=14\), so \(x=7\). Then \(7+y=10\), \(y=3\).

Worked example 2

\(3x+2y=12\) and \(5x-3y=1\).

The coefficients of \(y\) are 2 and \(-3\), so multiply the first by 3 and the second by 2 to get \(9x+6y=36\) and \(10x-6y=2\). Add: \(19x=38\), so \(x=2\). Substitute: \(6+2y=12\), \(y=3\). Check in the other equation: \(10-9=1\). Correct.

Worked example 3

\(4x+3y=18\) and \(2x+5y=16\).

Multiply the second by 2: \(4x+10y=32\). Subtract: \(7y=14\), \(y=2\). Then \(4x+6=18\), \(x=3\).

Worked example 4

\(x^2+y^2=25\) and \(y=x+1\).

\(x^2+(x+1)^2=25\) gives \(2x^2+2x-24=0\), so \(x^2+x-12=0\), \((x+4)(x-3)=0\). \(x=3\) gives \(y=4\); \(x=-4\) gives \(y=-3\). The solutions are \((3,4)\) and \((-4,-3)\). Geometrically, the line cuts the circle at two points.

Worked example 5

\(y=x^2-5\) and \(y=x+1\): \(x^2-x-6=0\), \(x=3\) or \(-2\), giving \((3,4)\) and \((-2,-1)\).

Worked example 6

A café sells 2 teas and 3 coffees for £9.00 and 3 teas and 1 coffee for £6.50. Find the price of a tea and of a coffee [4 marks].

\(2t+3c=9\), \(3t+c=6.5\). Multiply the second by 3: \(9t+3c=19.5\). Subtract the first: \(7t=10.5\), so \(t=1.5\). Then \(c=6.5-4.5=2\). A tea costs £1.50 and a coffee £2.00. Check in the first equation: \(3+6=9\).

Worked example 7

Solve \(2x+5y=1\) and \(3x-2y=11\) [4 marks].

Multiply the first by 2 and the second by 5: \(4x+10y=2\) and \(15x-10y=55\). Add: \(19x=57\), \(x=3\). Then \(6+5y=1\), \(y=-1\).

Worked example 8 (Higher tier)

Solve \(y=2x+1\) and \(y=x^2-2\) [4 marks].

\(x^2-2=2x+1\), so \(x^2-2x-3=0\), \((x-3)(x+1)=0\). \(x=3\), \(y=7\) or \(x=-1\), \(y=-1\).

Worked example 9: the sign rule for elimination

Solve \(5x+3y=19\) and \(2x+3y=10\) [3 marks].

The \(y\) coefficients are the same, with the same sign, so subtract the second equation from the first: \(3x=9\), so \(x=3\). Substitute into the second: \(6+3y=10\), \(3y=4\), \(y=\dfrac43\). Check in the first: \(15+4=19\) ✓. If the signs had been different, such as \(+3y\) and \(-3y\), you would have added.

Worked example 10: a worded problem with fractions

Three pens and two rulers cost £4.40. Two pens and five rulers cost £5.50. Find the cost of each [4 marks].

Equations: \(3p+2r=4.40\) and \(2p+5r=5.50\). Multiply the first by 2 and the second by 3: \(6p+4r=8.80\) and \(6p+15r=16.50\). Subtract: \(11r=7.70\), so \(r=0.70\). Then \(3p+1.40=4.40\), so \(p=1.00\). A pen costs £1.00 and a ruler 70p. Check: \(2+3.50=5.50\) ✓.

Worked example 11: graphical solution and meaning

Draw \(y=x+2\) and \(y=8-2x\) and find the solution [3 marks].

Table for \(y=x+2\): \((0,2)\), \((2,4)\), \((4,6)\). Table for \(y=8-2x\): \((0,8)\), \((2,4)\), \((4,0)\). Both lines pass through \((2,4)\), so the solution is \(x=2\), \(y=4\). Check algebraically: \(x+2=8-2x\) gives \(3x=6\), \(x=2\). This is a good method of checking graph answers.

Worked example 12: a break-even problem

A baker’s costs are \(C=150+0.8n\) and her income is \(I=2.5n\) for \(n\) loaves. How many loaves must she sell to break even? [3 marks]

Break-even is when \(C=I\): \(150+0.8n=2.5n\), so \(150=1.7n\) and \(n=88.2\). She must sell 89 loaves to make a profit, because she cannot sell 0.2 of a loaf. Check: 89 loaves give income \(222.50\) and costs \(221.20\), a profit of £1.30.