Top answers

How do I solve simultaneous equations?

A easy way to solve simultaneous equations is to substitiute one equation into another and then equate to find one of the unknowns. Once this has been found its a simple matter of substituting it back into t...
KL
Answered by Kishen L. • Maths tutor
3889 Views

Frank, Mary and Seth shared some sweets in the ratio 4:5:7. Seth got 18 more sweets than Frank. Work out the total number of sweets they shared.

The ratio 4:5:7 is a way of comparing the size of the shares received by Frank, Mary and Seth. It helps to pretend that we are sharing the sweets between some equally-sized boxes: each box has the same numbe...
RN
Answered by Richard N. • Maths tutor
30805 Views

Solve the simultaneous equations “x^2+y^2=4” and “x=2-y”. What does this tell us about the circle centred on the origin, with radius 2, and the straight line with y-intercept 2 and gradient -1?

In order to solve the pair of simultaneous equations, we must find a single set of values for x and y which fulfils both at once. By substituting “x=2-y” into “x 2 +y 2 =4”, we obtain a single equation conta...
AD
Answered by Aaron D. • Maths tutor
12732 Views

What is the length of the hypotenuse in the right angled triangle to one decimal place? (a=5cm, b=4cm)

Step 1. Use the formula a 2 +b 2 =c 2 Step 2. Square a and b to find a 2= 25 and b 2 =16 Step 3. Add a 2 and b 2 to find c 2 =41 Step 4. Find the square root of c 2 is sqrt(41)=6.403 Step 5. Round 6.403 to 6...
ER
Answered by Eleanor R. • Maths tutor
4412 Views

Prove the quadratic formula for ax^2 + bx + c = 0, where a is non 0 and a,b and c are reals.

By completing the square: ax^2 + bx + c = 0 => x^2 + (bx)/a + c/a = 0 (divide both side by a, since a is non-zero) => (x + b/(2a))^2 + c/a - (b/(2a))^2 = 0 (If this is not immediately clear, try expand...
SN
Answered by ShenZhen N. • Maths tutor
10013 Views