Hi, I have been fine with all the simultaneous equations I have done so far, how

Question

Hi, I have been fine with all the simultaneous equations I have done so far, howerever these questions I keep getting the wrongs answers to:

a) 4x= y +7
3x^2 - 4y = 7

b) 3x + 2y = 4
y^2 +2xy = 0

c) 5x+3y=4
5x^2 -3y^2 = 8

I know you have to substitute for these ones, and that you have to rearrange the first equation to x or y=.

However, sometimes I get algebraic fractions like y+7/4, and when I substitute in I find it really hard to make the correct quadratic equation to factorise to find x, and then eventually y by substitution into the original equation.

Please help, this would be much appreciated, thanks. :)

Explanation / Answer

Pay extra attention to the fractions. Multiply both sides of the equation by the denominator, to get rid of it.

Let's solve these systems.

a) 4x= y +7
3x^2 - 4y = 7

y = 4x - 7
3x^2 - 4(4x - 7) = 7
3x^2 - 16x + 28 = 7
3x^2 - 16x + 21 = 0

x = (16 +/- sqrt(256 - 4*3*21)) / 6 =
(16 +/- sqrt(4)) / 6 =
3 or 7/3

Substituting back on 4x = y +7:
12 = y + 7 --> y = 5
4*7/3 = y + 7 --> y = 7/3

Solutions: (3, 5) and (7/3, 7/3). Check them back in the original equations.

b) 3x + 2y = 4
y^2 +2xy = 0

At once, the second equation can be simplified to
y(y + 2x) = 0

So, either y = 0, or y + 2x = 0 --> y = -2x.

For y = 0, the first equation becomes 3x = 4 --> x = 4/3
For y = -2x. the first equation becomes 3x - 4x = 4 --> x = -4. So y = (-2)*(-4) = 8.

Solutions: (4/3, 0) and (-4, 8). Check them back in the original equations.

c) 5x+3y=4
5x^2 -3y^2 = 8

3y = 4 - 5x, or y = (4 - 5x) / 3. Substituting:

5x^2 - 3 * ((4 - 5x) / 3)^2 = 8
5x^2 - 3 * (4 - 5x)^2 / 9 = 8
5x^2 - (4 - 5x)^2 / 3 = 8
Multiplying both sides by 3:
15x^2 - (4 - 5x)^2 = 24
15x^2 - (16 - 40x + 25x^2) = 24
15x^2 - 16 + 40x - 25x^2 = 24
-10x^2 + 40x - 40 = 0
-x^2 + 4x - 4 = 0
x^2 - 4x + 4 = 0
(x - 2)^2 = 0
x = 2 (double root)

From that, 5x + 3y = 4 --> 10 + 3y = 4, or y = -2.

Solution: (2, -2)

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