Algebra

QUESTION 1 [2 marks] Easy

Solve for $x$: $4x-7=21$.

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$4x=21+7=28$, so $x=\dfrac{28}{4}=7$.

QUESTION 2 [2 marks] Easy

Solve for $x$: $3(x+5)=2x+21$.

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Distribute: $3x+15=2x+21$. Subtract $2x$ from both sides: $x+15=21$, so $x=6$.

QUESTION 3 [3 marks] Medium

Solve for $x$: $\dfrac{2x-3}{5}=\dfrac{x+4}{3}$.

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Cross-multiply: $3(2x-3)=5(x+4)$, so $6x-9=5x+20$. Subtract $5x$: $x-9=20$, so $x=29$.

QUESTION 4 [3 marks] Medium

If $5x-2=3x+12$, what is the value of $2x-1$?

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$5x-2=3x+12 \Rightarrow 2x=14 \Rightarrow x=7$. Then $2x-1=2(7)-1=13$. (Notice $2x-1$ can be evaluated directly once $2x=14$ is known, without solving for $x$ first.)

QUESTION 5 [4 marks] Hard

For what value of $k$ does the equation $6x+9=k(2x+3)$ have infinitely many solutions?

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Expand the right side: $6x+9=2kx+3k$. For every value of $x$ to be a solution, the coefficients of $x$ and the constant terms must match on both sides: $2k=6$ and $3k=9$. Both give $k=3$, so $k=3$ is the answer (with $k=3$ the equation becomes $6x+9=6x+9$, true for all $x$).

QUESTION 6 [5 marks] Hard

How many solutions does the equation $\dfrac{1}{2}(4x-6)-3=\dfrac{1}{3}(6x-9)$ have?

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Simplify the left side: $\dfrac{1}{2}(4x-6)-3=2x-3-3=2x-6$. Simplify the right side: $\dfrac{1}{3}(6x-9)=2x-3$. The equation becomes $2x-6=2x-3$. Subtracting $2x$ from both sides gives $-6=-3$, which is false no matter what $x$ is — so the equation has no solution (0 solutions). The $x$-terms cancel out entirely, which is the key thing to notice before grinding through more algebra.

QUESTION 7 [5 marks] Hard

If $ax-7=3x+b$ is true for every value of $x$, where $a$ and $b$ are constants, what is the value of $a-b$?

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For a linear equation to hold for every value of $x$, both sides must be identical: the coefficients of $x$ must match ($a=3$) and the constant terms must match ($-7=b$, so $b=-7$). Then $a-b=3-(-7)=10$.

Linear equations in one variable 7 questions