Question 182273
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Put them both in standard form, namely *[tex \Large Ax + By = C]


*[tex \LARGE \text{(1)          }\math 6x - 3y = 24]


*[tex \LARGE \text{(2)          }\math 10x - y = 7]


Multiply equation (2) by -3:


*[tex \LARGE \text{(3)          }\math -30x + 3y = -21]


Add Eq (3) to Eq (1):


*[tex \LARGE \text{             }\math -24x + 0y = 3\ \ \Rightarrow\ \ x = -\frac{1}{8}]


Substitute *[tex \Large x = -\frac{1}{8} ] into Eq (1)


*[tex \LARGE \text{             }\math 6\left(-\frac{1}{8}\right) - 3y = 24]


I'll let you do the arithmetic to determine the value of <i>y</i>.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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