Theresatuagnonidn
Answered

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how do i solve this 2 linear equation??
1. 2x-3y=5
   -4x+6y=-10

2. 2x+8y=3
   -4x+16y=-8

Sagot :

Kaiusidn
1. *multiply the first equation by two, to cancel the X values: 2(2x - 3y = 5); 4x - 12y = 10; *then add to the second equation: (4x - 12y = 20) + (-4x + 6y = -10) = (-6y = 10) = (y = -5/3); *substitute the value of y in the first equation to get x: [2x - 3(-5/3) = 5] = [2x + 5 = 5] = [x = 0]; 2. *multiply the first equation by two, to cancel the X values: 2(2x + 8y = 3); 4x + 16y = 6; *then add to the second equation: (4x + 16y = 6) + (-4x + 16y = -8) = (32y = -2) = (y = -1/16); *substitute the value of y in the first equation to get x: [2x - 16(-1/16) = 3] = [2x + 1 = 3] = [x = 1];
Riza1idn
[tex]method \ of \ elimination \\\\1)\\\\\begin{cases} 2x-3y=5\ \ /\cdot 2 \\ -4x+6y=-10 \end{cases}\\\\\begin{cases} 4x-6y=10 \\ -4x+6y=-10 \end{cases}\\+---------\\0=0 \\ \\ Answer: \ infinitely \ solutions [/tex]


[tex]2) \\\\\begin{cases} 2x+8y=3\ \ /\cdot 2 \\ -4x+16y=-8 \end{cases}\\\\\begin{cases}4x+16y=6\\ -4x+16y=-8 \end{cases}\\+--------\\32y=-2\ \ /:32\\y=-\frac{2}{32}=-\frac{1}{16}[/tex]

[tex] 2x+8y=3 \\ 2x+8 \cdot (-\frac{1}{16})=3 \\2x-\frac{1}{2}=3\\2x=3+\frac{1}{2}\\ \\2x=\frac{6}{2}+\frac{1}{2}\\ \\2x=\frac{7}{2}\ \ \cdot (\frac{1}{2})\\ \\x=\frac{7}{4}\\\\Answer: \ x=\frac{7}{4} \ \ and \ \ y= -\frac{1}{16} [/tex]