Question

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Answer 1

Answer:

$\huge\underline\mathfrak\red{❥Answer}$

Step-by-step explanation:

$x + y = 6 \\ x + y - y = 6 - y \\ x + 0 = 6 - y \\ x = 6 - y \\substitute \: (6 - y) \: for \: x \: in \: the \: first \\ eq \: and \: slove \: y \\ {x}^{2} + {y}^{2} = 20 \\ {(6 - y)}^{2} + {y}^{2} = 20 \\ 36 - 12y + {y}^{2} + {y}^{2} = 20 \\ {1y}^{2} + {1y}^{2} - 12y + 36 = 20 \\ (1 + 1) {y}^{2} - 12y + 36 = 20 \\ {2y}^{2} - 12y + 36 = 20 \\ {2y}^{2} - 12y + 36 - 20 = 20 - 20 \\ (2y - 8)(y - 2) = 0 \\ solution \: y \\ 2y - 8 = 0 \\ 2y - 8 + 8 = 0 + 8 \\ 2y = 8 \\ \frac{2y}{2} = \frac{8}{2} \\ y = 8 \: \\ solution \: 2 \: for \: y \\ y - 2 = 0 \\ y - 2 + 2 = 0 + 2 \\ y = 0 Substitute the values for y into \\ the solution to the second \\ equation at the end \\ of Step 1 and calculate x: \\ x = 2 \: y = 4$