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QUESTION IMAGE

match each expression with the number of solutions the equation would h…

Question

match each expression with the number of solutions the equation would have if that expression completed the equation on the right side.

6(2x − 3) \t\ta. no solution

4(3x − 3) \t\tb. one solution

4(2x − 3) \t\tc. all solutions

Explanation:

Step1: Analyze \(6(2x - 3)\)

Assume the left - hand side of the equation is a linear expression (usually in the form \(ax + b\), for example, if the left - hand side is \(12x-18\) and we set the equation as \(12x - c=6(2x - 3)=12x-18\), if the left - hand side is \(12x + d\) (where \(d
eq - 18\)), then \(12x + d=12x-18\), subtracting \(12x\) from both sides gives \(d=-18\), which is a contradiction if \(d
eq - 18\). But if we assume the left - hand side is also a linear expression. Wait, actually, when we have an equation like \(a(2x - 3)=6(2x - 3)\). If \(a = 6\), then the equation is \(6(2x - 3)=6(2x - 3)\), which is an identity (true for all \(x\)). But if we consider the general case, when we expand \(6(2x - 3)=12x-18\). If the left - hand side of the equation (the part we are not shown) is, say, \(12x + k\) where \(k
eq - 18\), then \(12x + k=12x-18\) has no solution. But if the left - hand side is \(12x-18\), then it has all solutions. Wait, maybe we assume the left - hand side is a linear expression of the form \(m(2x - 3)\) or something else. Wait, actually, let's consider the three expressions:

First, let's expand each expression:

  • \(6(2x - 3)=12x-18\)
  • \(4(3x - 3)=12x-12\)
  • \(4(2x - 3)=8x-12\)

Now, for a linear equation of the form \(Ax + B = Cx+D\):

  • If \(A = C\) and \(B = D\), the equation has all solutions (identity).
  • If \(A = C\) and \(B

eq D\), the equation has no solution (contradiction).

  • If \(A

eq C\), the equation has one solution (we can solve for \(x\) as \(x=\frac{D - B}{A - C}\))

Now, let's assume that the left - hand side of the equation (the part we are not given) is of the form \(12x + k\) (for the first two expressions which have \(x\) coefficient 12) and \(8x + m\) for the third.

  1. For \(6(2x - 3)=12x-18\):
  • If the left - hand side of the equation (the unshown part) is also \(12x-18\), then the equation \(12x-18 = 12x-18\) has all solutions (c). But if the left - hand side is \(12x + k\) where \(k

eq - 18\), it has no solution. But maybe the left - hand side is of the form \(a(2x - 3)\). Wait, maybe the original equation is of the form \(n(2x - 3)=6(2x - 3)\). If \(n = 6\), all solutions; if \(n
eq6\), then \(n(2x - 3)-6(2x - 3)=0\Rightarrow(n - 6)(2x - 3)=0\). If \(n
eq6\), then \(2x-3 = 0\Rightarrow x=\frac{3}{2}\), one solution. Wait, this is getting confusing. Maybe the intended approach is:

Let's consider the three expressions and match them to the number of solutions:

  • Expression 1: \(6(2x - 3)=12x-18\)
  • Expression 2: \(4(3x - 3)=12x-12\)
  • Expression 3: \(4(2x - 3)=8x-12\)

Now, suppose the left - hand side of the equation is \(12x-18\) (for the first expression). Then the equation \(12x-18=12x-18\) has all solutions (c).

For the second expression \(4(3x - 3)=12x-12\). If the left - hand side of the equation is \(12x-18\), then \(12x-18 = 12x-12\), subtracting \(12x\) from both sides gives \(-18=-12\), which is false (no solution, a).

For the third expression \(4(2x - 3)=8x-12\). If the left - hand side of the equation is \(12x-18\), then \(12x-18=8x-12\). Subtract \(8x\) from both sides: \(4x-18=-12\). Add 18 to both sides: \(4x = 6\), \(x=\frac{6}{4}=\frac{3}{2}\) (one solution, b).

Wait, maybe the standard way is:

  • When the two sides of the equation are equivalent (same coefficient of \(x\) and same constant term), all solutions.
  • When the coefficients of \(x\) are the same but constant terms are different, no solution.
  • When the coefficients of \(x\) are different, one solution.

So:

  1. \(6(2x - 3)=12x-18\). If we have an equation like \(12x-18 = 12x-18\)…

Answer:

  • \(6(2x - 3)\): c. all solutions
  • \(4(3x - 3)\): a. no solution
  • \(4(2x - 3)\): b. one solution