Simultaneous equations

With substitution, you rewrite one variable using the other equation, then swap it into the second equation.

Here, the first equation already gives $x$ in terms of $y$: $x=3y-1$

So every time $x$ appears in the other equation, we can swap in $(3y-1)$. This is a key skill for Systems of linear equations in Coordinate Geometry and Linear Modelling.

Start from the second equation: $2x+y=12$

Now replace $x$ with the entire expression $(3y-1)$.

Key point: the brackets matter because the $2$ must multiply all of $(3y-1)$, not just part of it.

Option B doesn’t actually replace $x$ inside $2x$; it keeps $2x$ and then adds $(3y-1)$. That leaves both $x$ and $y$ in the same equation, so it doesn’t move the substitution method forward.

Option D removes the brackets in a way that makes the $2$ multiply only the $3y$ term. The $-1$ is left outside, which breaks the distributive property because $2$ should apply to the entire expression for $x$.

Option C looks like it is trying to expand straight away, but it expands incorrectly. Also, the question asks for the first step after substitution, which should clearly show the replacement before doing extra algebra.

Replacing $x$ in $2x+y=12$ with $(3y-1)$ gives: $2(3y-1)+y=12$

That matches Option A.

When using substitution, you take the expression for one variable and plug it into the other equation.

We have:

1. $x=3y-1$
2. $2x+y=12$

Because the first equation tells us what $x$ equals, we should replace the $x$ in the second equation with $3y-1$.

Check whether each option actually replaces $x$.

In Option B, $(3y-1)$ gets placed as if it were replacing $y$, giving $2x+(3y-1)=12$. Since the equation we’re given is for $x$, swapping out $y$ is the wrong structural move.

If you substitute a multi-term expression like $3y-1$ into something like $2x$, you must keep parentheses so the coefficient multiplies every term.

Option D: $2(3y)-1+y=12$ Here the $2$ only affects $3y$, and the $-1$ is not multiplied by $2$. That’s not a correct substitution into $2x$.

Option C: $6y-1+y=12$ This tries to expand immediately, but it makes a typical mistake: it multiplies $2\times 3y$ to get $6y$, but it doesn’t multiply $2\times -1$.

Correct distribution would be: $2(3y-1)=6y-2$

Since Option C has $-1$ instead of $-2$, the expansion is wrong.

Option A correctly replaces only $x$ and keeps the brackets so the structure stays correct:

$2(\text{expression for }x)+y=12$ $2(3y-1)+y=12$

To confirm which equation could be the correct first substitution step, we can solve the system and see which option stays true.

Given: $\{\begin{array}{l}x=3y-1\\ 2x+y=12\end{array}$

We substitute the expression for $x$ into the second equation and solve for $y$.

Solve using substitution:

$\begin{array}{rl}2(3y-1)+y& =12\\ 6y-2+y& =12\\ 7y-2& =12\\ 7y& =14\\ y& =2\end{array}$

Then put $y=2$ into the first equation: $x=3(2)-1=5$

So the solution is $x=5$ and $y=2$.

Substitute $x=5$ and $y=2$ into Option B: $2x+(3y-1)=12$ $2(5)+(3(2)-1)=10+5=15$

Because $15\ne 12$, Option B cannot be correct.

Now check Option C: $6y-1+y=12$ $6(2)-1+2=12-1+2=13$

Since $13\ne 12$, Option C is not valid. It reflects an incorrect expansion from the substitution step.

Now check Option D: $2(3y)-1+y=12$ $2(3\cdot 2)-1+2=2(6)-1+2=13$

Since $13\ne 12$, Option D is also incorrect. It failed to multiply the entire substituted expression by $2$.

Finally, test Option A: $2(3y-1)+y=12$ $2(3(2)-1)+2=2(5)+2=12$

This is true ($12=12$), so Option A matches the correct substitution of $x=3y-1$ into $2x+y=12$.

We are working with a system of two linear equations:

1. $5x-2y=10$
2. $10x-4y=20$

To decide how many solutions there are, we can check whether one equation is just a scaled version of the other (which would mean they represent the same line).

The second equation looks like it might be a multiple of the first. Multiply every term in $5x-2y=10$ by $2$:

$\begin{array}{rl}2\times (5x-2y)& =2\times (10)\\ 10x-4y& =20\end{array}$

After scaling, we get $10x-4y=20$, which is exactly the same as the second equation.

So both equations describe the same line. That means every point on that line satisfies both equations.

Because the two equations are the same line, we can eliminate:

• Option A: This would happen if the lines crossed once (different gradients).
• Option C: A pair of straight-line equations cannot have exactly two solutions.
• Option D: This would happen if the lines were parallel but different (same gradient, different intercept).

Since both equations represent the same line, there are infinitely many ordered pairs $(x,y)$ that satisfy both equations.

We are given the system:

$\{\begin{array}{l}5x-2y=10\text{— (Equation 1)}\\ 10x-4y=20\text{— (Equation 2)}\end{array}$

We want to determine whether the system has one solution, no solution, or infinitely many solutions. We will use the Algebraic Elimination method, which is covered in Coordinate Geometry and Linear Modelling.

To eliminate a variable, we want matching coefficients. Since Equation 2 has $10x$, multiply Equation 1 by $2$:

$2\times (5x-2y=10)$ $10x-4y=20$

Subtract the scaled Equation 1 from Equation 2:

$\begin{array}{rl}(10x-4y)-(10x-4y)& =20-20\\ 0x+0y& =0\\ 0& =0\end{array}$

Both $x$ and $y$ disappear at the same time.

The result $0=0$ is a mathematical identity, meaning it is true for all values of $x$ and $y$.

For a linear system, this shows the equations are dependent: they are just different-looking versions of the same line. Any $(x,y)$ that works in Equation 1 will also work in Equation 2.

If you graph both equations on a GDC, you will find the graphs overlap completely, showing they are the same line.

Consider the answer choices:

• Option A: Incorrect. One solution happens when two lines intersect once (different slopes).
• Option C: Incorrect. Two linear equations cannot produce exactly two solutions.
• Option D: Incorrect. No solution would give a contradiction such as $0=5$, meaning parallel distinct lines.

Because elimination produced the identity $0=0$, the system has infinitely many solutions.

To determine the number of solutions, we can rewrite both equations in slope-intercept form:

$y=mx+c$

Here, $m$ is the gradient and $c$ is the y-intercept. If two equations have the same $m$ and the same $c$, they represent the same line.

Isolate $y$ in the first equation:

$\begin{array}{rl}5x-2y& =10\\ -2y& =-5x+10\\ y& =\frac{-5}{-2}x+\frac{10}{-2}\\ y& =\frac{5}{2}x-5\end{array}$

So the gradient is $m=\frac{5}{2}$ and the y-intercept is $c=-5$.

Now rearrange the second equation:

$\begin{array}{rl}10x-4y& =20\\ -4y& =-10x+20\\ y& =\frac{-10}{-4}x+\frac{20}{-4}\\ y& =\frac{5}{2}x-5\end{array}$

Note that $\frac{10}{4}$ simplifies to $\frac{5}{2}$, and $20$ divided by $-4$ is $-5$.

A quick check on a graphing calculator will show both equations draw the same line, one lying directly on top of the other.

Since both equations simplify to $y=\frac{5}{2}x-5$, we eliminate:

• Option A: Different slopes would be needed for one intersection point.
• Option C: Straight lines cannot intersect in exactly two points.
• Option D: No solution would require equal slopes but different y-intercepts.

Because the two equations are the same line, they share infinitely many points in common, so the system has infinitely many solutions.

To represent the total cost ($y$), we use the Direct Algebraic Modeling idea: total cost equals the fixed cost plus the variable cost for each item.

From the question, we know:

• The fixed costs are $200$
• The variable cost is $5$ per item produced ($x$)

Setting up the equation: $\text{Total Cost}=(\text{Variable Cost}\times x)+\text{Fixed Cost}$ $y=5x+200$

Now we model the money earned from sales. Revenue is the amount collected from selling items.

The question states they sell each item for $10$. So revenue ($y$) is the selling price times the number sold ($x$):

$\text{Revenue}=\text{Price per item}\times x$ $y=10x$

Next, we compare our equations ($y=5x+200$ and $y=10x$) to the answer choices and remove the ones that disagree:

• Option A: Uses $y=5x-200$. Fixed costs increase total expenses, so they should be added, not subtracted.
• Option B: Uses $y=10x+200$ for revenue. Revenue comes only from sales, so the $+200$ does not belong there.
• Option C: Uses $y=200x+5$. This mixes up the roles of the numbers, treating $200$ like the per-item rate and $5$ like the fixed amount.

Break-even happens when the total cost equals the total revenue. Using a calculator or graphing tool, you can see where the two lines intersect to confirm the system represents that situation.

So the correct system that models total cost and total revenue is $y=5x+200$ and $y=10x$.

In linear modelling, a cost rule is often written as $y=mx+c$.

• The fixed cost is the part you pay even when $x=0$, so it is the $y$-intercept ($c$).
• The variable cost is the amount per item, so it is the slope ($m$).

From the question, we identify:

• Fixed cost: $200$
• Variable cost: $5$ per item

So the cost equation must be: $y=5x+200$

Revenue is the money earned from selling items. If you sell $0$ items, you earn $0$, so there is no constant term; it follows $y=mx$.

• Selling price: $10$ per item

So, the revenue equation must be: $y=10x$

Now eliminate options that clearly misuse the fixed cost or swap the roles of the numbers:

• Option A is wrong because it uses $-200$. A fixed cost adds to expenses; it doesn’t reduce them.
• Option C is wrong because it reverses the variable and fixed parts ($200x+5$ would mean $200$ per item, which is not stated).

Look at what is left and focus on where the $+200$ appears.

Option B claims cost is $y=5x$ and revenue is $y=10x+200$. That incorrectly places the fixed cost in the revenue equation. The fixed cost must be part of production cost, so it belongs in $y=5x+200$.

The only option that correctly uses $5$ as the cost rate, $200$ as the fixed cost, and $10$ as the selling price is Option D.