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# Balancing Scales

What is the weight of one square?

In the previous quiz you used the method of elimination to help solve multivariable equations. Another strategy to use is **substitution** in which we replace one variable with another.

We are given the following information and want to determine how many squares equal a triangle.

Using substitution, we can replace each of the circles in the first equation with two triangles, since we know that 1 circle = 2 triangles.

Now we have an equation that gives us the relationship between triangles and squares. 4 triangles = 1 square.

What is the combined weight of a triangle and a circle?

If we wanted to use the substitution method, what would be the next best step in solving for this system of equations? \[\begin{align}2x - 3y &= 9 \\ x &= 4 + y\end{align}\]

**A.** Solve for \(y\) in the bottom equation.

**B.** Using substitution, replace \(x\) in the top equation with \(4 + y.\)

**C.** Add the top and bottom equations together.

**D.** Divide by 2 and then solve for \(x\) in the top equation.

One way to use **substitution** to solve two variable equations is to solve both equations for one value and then use substitution to set the two results equal to each other.

It's a bit tricky, so here is an example: \[\begin{align} 3s + t &= 16 \\ 2s + 2t &= 10.\end{align}\] In this system, \(t\) is going to be the easiest variable to solve for because solving for \(t\) will not create any fractions.

Solving the first equation for \(t,\) we get \(t = 16 - 3s.\)

Solving the second equation for \(t,\) we get \(t = 5 - s.\)

Because we know that \(t=t,\) we can solve the system by setting \(16 - 3s = 5 - s\) and then solving.

We can express the equation \(2s + 2t = 16\) in terms of \(s\) by solving for \(t:\) \[\begin{align} 2s + 2t &= 16 \\ 2t &= 16 - 2s \\ t &= 8 - s\,.\end{align}\]

How we can we express \(t\) in terms of \(s\) for the equation below? \[3t - 4s = 3 \]

\[\left\{\begin{array}{cl} t &= 8-s \\\\ t&= \frac{4}{3}s + 1 \end{array}\right. \]

These are the two equations from the previous problem, both solved for \(t\) in terms of \(s.\) Using substitution, we can substitute \(8-s\) for \(t\) in the second equation: \[8-s = \frac{4}{3}s+1.\]

What is \(s\,?\)

Solve for \(z:\) \[\left\{\begin{array}{cl} 4w + 2z &= 0 \\ 3w + z&= -2. \end{array}\right. \]

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