QUESTION IMAGE
Question
solve the following system of equations for z and for y:
system of equations: value of z value of y
y = 9 + 3z
y = 39 - 3z
solve the following system of equations for a and for b:
system of equations: value of a value of b
4a + 2b = 20
15a + 5b = 60
plot the following system of equations on the following graph.
system of equations:
p = 6 - 2q
p = 4 + q
note: use the orange line (square symbols) to plot the first equation, and use the blue line (circle symbols) to plot the second equation.
Solving for \(z\) and \(y\)
Step1: Set the two - equations equal
Since \(y = 9 + 3z\) and \(y=39 - 3z\), we set \(9 + 3z=39 - 3z\).
Step2: Solve for \(z\)
Add \(3z\) to both sides: \(9+3z + 3z=39-3z + 3z\), which simplifies to \(9 + 6z=39\). Then subtract 9 from both sides: \(6z=39 - 9=30\). Divide both sides by 6: \(z=\frac{30}{6}=5\).
Step3: Solve for \(y\)
Substitute \(z = 5\) into \(y = 9+3z\), so \(y=9 + 3\times5=9 + 15=24\).
Step1: Simplify the first equation
Divide the equation \(4a + 2b=20\) by 2 to get \(2a + b = 10\), which can be rewritten as \(b=10 - 2a\).
Step2: Substitute \(b\) into the second equation
Substitute \(b = 10 - 2a\) into \(15a+5b = 60\). We have \(15a+5(10 - 2a)=60\). Expand: \(15a+50 - 10a=60\). Combine like - terms: \(5a+50 = 60\). Subtract 50 from both sides: \(5a=60 - 50 = 10\). Divide both sides by 5: \(a = 2\).
Step3: Solve for \(b\)
Substitute \(a = 2\) into \(b=10 - 2a\), so \(b=10-2\times2=10 - 4 = 6\).
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Value of \(z\): \(5\)
Value of \(y\): \(24\)