Question

for integers x and y, the inequality is satisfied, find the largest value of the x+y.
$3 \sqrt{4x - 5y + 7} + 5 \div 3x - 4y + 61 \leqslant 4$
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Answer 1

$\frac{3 \sqrt{4x - 5y + 7} + 5}{3x - 4y + 61} \leqslant 4 \\ {3 \sqrt{4x - 5y + 7} + 5} \leqslant 4 \times (3x - 4y + 61) \\ {3 \sqrt{4x - 5y + 7} + 5} \leqslant 12x - 16y + 244 \\3 \sqrt{4x - 5y + 7} \leqslant12x - 16y + 244 - 5 \\ 3 \sqrt{4x - 5y + 7} \leqslant 12x - 16y + 239 \\ on \: squaring \: both \: sides \\ {(3 \sqrt{4x - 5y + 7}) }^{2} \leqslant {(12x - 16y + 239)}^{2} \\ 9 \times (4x - 5y + 7) \leqslant 144 {x}^{2} + 256 {y}^{2} + 57121 - 384xy - 7648y + 5736x \\ 36x - 45y + 63 + 7648y - 5736x - 57121 \leqslant 144 {x}^{2} + 256 {y}^{2} - 384xy \\ 7603y - 5700x - 57058 \leqslant {(12x - 16y)}^{2} \\ \\ \\ i \: think \: there \: is \: a \: mistake \: in \: the \: question.$

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