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Question
a) \\( x < 4, \\ y < 3, \\ -x + y \leq 1 \\)
b) \\( x < 3, \\ y < 4, \\ -x + y \leq 1 \\)
c) \\( x < 4, \\ y < 3, \\ x + y \leq 1 \\)
d) \\( x < 3, \\ y < 4, \\ x + y \leq 1 \\)
e) \\( x < 4, \\ y < 3, \\ x - y \leq 1 \\)
f) none of the above.
question 5
your answer is correct.
solve the system for \\( x \\).
\\( 12x + 6y = 36 \\)
\\( x + y = 3 \\)
) \\( x = 3 \\)
) \\( x = -3 \\)
\\( y = 4 \\)
Step1: Simplify the second equation
From \(x + y = 3\), we can express \(y\) as \(y = 3 - x\).
Step2: Substitute \(y\) into the first equation
Substitute \(y = 3 - x\) into \(12x + 6y = 36\). We get \(12x + 6(3 - x)=36\).
Step3: Expand and simplify the equation
Expand the left - hand side: \(12x+18 - 6x = 36\). Combine like terms: \(6x+18 = 36\). Subtract 18 from both sides: \(6x=36 - 18=18\).
Step4: Solve for \(x\)
Divide both sides of the equation \(6x = 18\) by 6. We have \(x=\frac{18}{6}=3\).
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\(x = 3\) (assuming the correct option is the one with \(x = 3\), for example, if the option is like "A. \(x = 3\)", then the answer is A. \(x = 3\))