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3 years ago

Please help with another math question

Mathematics

1 answer:

Vladimir79 [104]3 years ago

8 0

Given system of equations:
2y=4x+20 Eq 1
4y=6x+24 Eq 2
Divide both sides of equation 1 by 2.
y=2x+10
Substitute value of y in equation 2.
4(2x+10)=6x+24
8x+40=6x+24
2x+40=24
2x=-16
x=-8
Put value of x in y=2x+10
y=2(-8) + 10
y=-16+10
y=-6

Answer : (-8,-6)

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$(x+4)\times(x-0)\times(x-1)\times(x-4) \\\\= x(x+4)(x-4)(x-1)\\\\= x(x^2-4^2)(x-1)\\\\= x(x^2-16)(x-1)\\\\= x(x^2-16)(x-1)\\\\=x(x^2x+x^2\left(-1\right)+\left(-16\right)x+\left(-16\right)\left(-1\right))\\\\= x(x^3-x^2-16x+16)\\\\=x^4-x^3-16x^2+16x$

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