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How do you solve simultaneous equations?

The easiest way to solve simultaneous equations is by elimination. This is the idea of cancelling out one of your variables, the X or Y so that you can solve the remaining variable and then substitute this a...
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Answered by Ben H. • Maths tutor
4199 Views

Expand (x+4)(x+3).

To answer this you multiply everything in the left bracket by everything in the right bracket, so rewrite the equation as x(x+3)+4(x+3). Then you can expand each more easily: x(x+3) = x 2 +3x 4(x+3) = 4x+4×3...
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Answered by Elliot D. • Maths tutor
32200 Views

Simplify: 2x +6y +2y - x

collect x terms together i.e.) 2x - X = 1x and then collect y terms together i.e) 6y + 2y = 8y add your two answers together to get: x + 8y Answer x +8y
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Answered by Sophie M. • Maths tutor
6861 Views

If a and b are the roots of the quadric polynomial 2x^2+6x+7 what are a+b and ab?

The phrase "a and b are roots of 2x^2+6x+7" is just a way of saying that x=a and x=b solve the equation 2x^2+6x+7=0. Check out that by diving by 2 on both sides of this equation we get that that x=...
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Answered by Guillermo C. • Maths tutor
3807 Views

A bag contains beads, 60% of which are green. A student claims that the probability of getting two green beads if the beads aren't replaced is 1/3 as 6/10 * 5/9 is 1/3. Is the student right?

They have multiplied the fractions correctly but they are still incorrect. The student has assumed that a bag with 60% green beads contains 10 beads. If the bag had 100 beads and 60 were green, 60% of the be...
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Answered by Adithya K. • Maths tutor
3897 Views