Solve \(2(m+3)=4\).
\(2(m+3)\)
\( = 4\)
\(2(m)+2(3)\)
\(=4\)
\(2m+6\)
\(2m\)
\(=4-6\)
\(=-2\)
\(m\)
\(= \dfrac{-2}{2}\)
\(=-1\)
\(m+3\)
\(= \dfrac{4}{2}\)
divide by \(2\)
\(=2\)
\(=2-3\)
subtract \(3\)
\(= -1\)
Solve \(7g=8+5g\) and check your answer.
Notice there is a variable, \(g\), on both sides of the equals sign.
\(7\class{timed add1-hl2 remove2-hl2}{g}\)
\(=8+5\class{timed add1-hl2 remove2-hl2}{g}\)
Goal: To isolate the variable
\(7g \class{hl2}{-5g}\)
\( = 8 + 5g \class{hl2}{ -5g}\)
\(2g\)
\(=8\)
\(g\)
\(=\dfrac{8}{2}\)
\(\text{LS}\)
\(=7g\)
\(=(7)(4)\)
\(=28\)
\(\text{RS}\)
\(=8+5g\)
\(=8+5(4)\)
\(=8+20\)
Since \(\text{LS}=\text{RS}\), \(g=4\) is the correct solution.
Solve \(6w+5=9w-10\).
\(6w+5\)
\( = 9w-10\)
\(5\)
\(=9w-10 \class{hl2}{-6w}\)
subtract \(6w\) from both sides
\(=3w-10\)
\(5\class{hl2}{+10}\)
\(=3w\)
add \(10\) to both sides
\(15\)
\(\dfrac{15}{\class{hl2}{3}}\)
\(=w\)
divide both sides by \(3\)
Since the coefficients of \(w\) are \(6\) and \(9\), leave the one with the higher coefficient \(9w\) on the right side and move \(6w\) from the left to the right.
Can you do a quick mental check to see if the left and right sides are equal?
Since the coefficients of \(w\) are \(6\) and \(9\), leave the one with the higher coefficient, \(9w\), on the right side and move \(6w\) from the left to the right.
\(=9w-10\)
\(5 \class{hl2}{+10}\)
\(=9w\class{hl2}{-6w}\)
move both \(6w\) and \(-10\) at the same time
\(\dfrac{15}{3}\)
Solve \(4(h-3)=15h+21\).
Before attempting to isolate \(h\), we will need to expand the left side to remove the brackets, since both sides are not divisible by \(4\).
\(\begin{align*} 4(h-3)&=15h+21\\ 4(h)+4(-3)&=15h+21\\ 4h-12&=15h+21 \end{align*}\)
Now this looks like a question that we have seen before, with both variable and constant terms on both sides of the equal sign. Use inverse operations in one step (as shown) or two steps to isolate \(h\).
\(\begin{align*} 4h-12&=15h+21\\ -12-21&=15h-4h\\ -33&=11h\\ \frac{-33}{11}&=h\\ -3&=h \end{align*}\)
Solve \(4(m-2)=5(m+4)\) and check your answer.
Multiply to remove the brackets, then use inverse operations to isolate \(m\).
\(\begin{align*} 4(m-2)&=5(m+4)\\ 4(m)+4(-2)&=5(m)+5(4)\\ 4m-8&=5m+20\\ -8-20&=5m-4m\\ -28&=m \end{align*}\)
Check
\(\begin{align*} \text{LS}&=4(m-2)\\ &=4(-28-2)\\ &=4(-30)\\ &=-120 \end{align*}\)
\(\begin{align*} \text{RS}&=5(m+4)\\ &=5(-28+4)\\ &=5(-24)\\ &=-120 \end{align*}\)
Since \(\text{LS}=\text{RS}\), \(m=-28\) is the correct solution.
An integer is multiplied by \(6\), and then it is subtracted from \(50\). The result is \(8\). What is the integer?
In order to find the integer, we want to use a method that does not involve trial and error.
To do this, we will need to complete the following: