SBMPTN Zone : Sistem Persamaan Linear Tiga Variabel

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Tipe:

Standar SBMPTN

No. 1

Jika 3x+y+z=k, x+3y+z=k, x+y+3z=k, k\neq0, maka x^2+y^2+z^2 jika dinyatakan dalam k adalah ....

1. \dfrac{k^2}{25}
2. \dfrac{3k^2}{25}
3. \dfrac{k^2}5

4. \dfrac{9^2}{16}k^2
5. \dfrac3{15}k^2

Ganesha Operation

\(\begin{aligned} 3x+y+z&=x+3y+z\\ x&=y \end{aligned}\)

\(\begin{aligned} x+3y+z&=x+y+3z\\ y&=z \end{aligned}\)

\(\begin{aligned} 3x+y+z&=k\\ 5x&=k\\ x&=\dfrac{k}5 \end{aligned}\)
x=y=z=\dfrac{k}5

\(\begin{aligned} x^2+y^2+z^2&=\left(\dfrac{k}5\right)^2+\left(\dfrac{k}5\right)^2+\left(\dfrac{k}5\right)^2\\ &=\dfrac{k^2}{25}+\dfrac{k^2}{25}+\dfrac{k^2}{25}\\ &=\boxed{\boxed{\dfrac{3k^2}{25}}} \end{aligned}\)

Jadi, x^2+y^2+z^2=dfrac{3k^2}{25}.
JAWAB: B

No. 2

Jika (a+b,a,b) memenuhi sistem persamaan:
3x-y+2z=-1
-2x+y+3z=-3
maka a+b=

1. 0
2. 1
3. 2

4. 3
5. 4

\(\begin{aligned} 3x-y+2z&=-1\\ 3(a+b)-a+2b&=-1\\ 3a+3b-a+2b&=-1\\ 2a+5b&=-1 \end{aligned}\)

\(\begin{aligned} -2x+y+3z&=-3\\ -2(a+b)+a+3b&=-3\\ -2a-2b+a+3b&=-3\\ -a+b&=-3\\ -2a+2b&=-6 \end{aligned}\)

\(\begin{aligned} 2a+5b&=-1\\ -2a+2b&=-6\qquad&+\\\hline 7b&=-7\\ b&=-1 \end{aligned}\)

\(\begin{aligned} -a+b&=-3\\ -a+(-1)&=-3\\ -a-1&=-3\\ -a&=-3+1\\ &=-2\\ a&=2 \end{aligned}\)

\(\begin{aligned} a+b&=2+(-1)\\ &=\boxed{\boxed{1}} \end{aligned}\)

Jadi, a+b=1.
JAWAB: B