Solve the following systems of equations. [math: 2x−3y=1] 2x-3y=1 [math: 3x+4y=−1] 3x+4y=-1

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First, I’ll rewrite the two equations clearly: 2x − 3y = 1 and 3x + 4y = −1. To compare the options, I’ll solve the system step by step.
Option a: x = −3/17 and y = −5/17. If we substitute these into 2x − 3y, we get 2(−3/17) − 3(−5/17) = −6/17 + 15/17 = 9/17, which does not equal 1. Therefore, option a is inconsistent with the first equation and is incorrect.
Option b: x = 1/17 and y = −5/17. Check in the first equation: 2x − 3y = 2(1/17) − 3(−5/17) = 2/17 + 15/17 = 17/17 = 1, which matches. Check in the second equation: 3x + 4y = 3(1/17) + 4(−5/17) = 3/17 − 20/17 = −17/17 = −1, which also matches. This option satisfies both equations, so it is a valid solution.
Option c: y = 117 and x = −5/17. Substituting y = 117 into the first equation gives 2x − 3(117) = 1, so 2x = 1 + 351 = 352, hence x = 176. This does not agree with x = −5/17, and also would distort the second equation. Therefore option c is incorrect.
Option d: x = 117 and y = 5/17. Let’s test in the first equation: 2x − 3y = 2(117) − 3(5/17) = 234 − 15/17, which is not equal to 1. The mismatch shows option d is false.
In summary, options a, c, and d fail to satisfy at least one of the equations, while option b passes both checks. The reason is that solving the system yields specific fractional values that satisfy both linear constraints, and any deviation from those values will violate at least one equation.

Solve the following systems of equations. [math: 2x−3y=1] 2x-3y=1 [math: 3x+4y=−1] 3x+4y=-1

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