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Sagot :
To solve the following simultaneous linear equation, we are going to solve for y on the first equation and replace it on the second as:
[tex]\begin{gathered} 8x+3y=-4 \\ 3y=-4-8x \\ y=\frac{-4-8x}{3} \\ y=\frac{-4}{3}-\frac{8x}{3} \end{gathered}[/tex]Replacing it on the second equation and solving for x, we get:
[tex]\begin{gathered} 5x+2y=6 \\ 5x+2(\frac{-4}{3}-\frac{8x}{3})=6 \\ 5x-\frac{8}{3}-\frac{16x}{3}=6 \\ 5x-\frac{16}{3}x=6+\frac{8}{3} \\ \frac{-1}{3}x=\frac{26}{3} \\ -x=26 \\ x=-26 \end{gathered}[/tex]Finally, replacing x on the first equation and solving for y, we get:
[tex]\begin{gathered} 8x+3y=-4 \\ 8(-26)+3y=-4 \\ 3y=-4+208 \\ 3y=204 \\ y=\frac{204}{3}=68 \end{gathered}[/tex]Answer: x = -26 and y = 68
To make subtraction of 5x - 16x/3, we can use the following equation:
[tex]\frac{a}{b}-\frac{c}{d}=\frac{a\cdot d-b\cdot c}{b\cdot d}[/tex][tex]\begin{gathered} 5x-\frac{16x}{3}=\frac{5x}{1}-\frac{16x}{3}=\frac{5x\cdot3-1\cdot16x}{1\cdot3}=\frac{15x-16x}{3}=\frac{-1x}{3} \\ \end{gathered}[/tex]
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